MHB Integration using Beta and Gamma Functions

[alyafey22]

Thank you very much @ZaidAlyafey, kudos to you, and I really appreciate this; the method is much clearer this way as I can understand better, and as I read carefully, all at once. So thank you ZaidAlyafey, I can't thank you enough! I owe you =)

The INTEGRAL:
----------------------
$$-\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$

$$B(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)} \,dt$$

Let's differentiate with respect to $b$

$$B_b(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)}\ln(1-t) \,dt$$

Let $t = x^2 \implies dt = 2x dx$

$$B_b(a,b) = (2) \int_{0}^{1} {x}^{2a-1}{(1-x^2)}^{(b-1)}\ln(1-x^2) \,dx$$

We need $b - 1 = 0 \implies b = 1$
We need $2a - 1 = -2 \implies a = -1/2$

$$B_b((-1/2),1) = (2) \int_{0}^{1} {x}^{-2}\ln(1-x^2) \,dx$$

Darn... We still can't do anything =(

You will run into a cycle. We want to find the integral without using the digamma function. Fortunately , there is an anti-derivative for the integral. Please , give it more thoughts.

The sum is $u = \psi(1/2)$ so

$$u = (-1/2)B(-1/2, 1)\gamma - (1/2)B(-1/2, 1)(u)$$

It is very surprising.

I couldn't compute it but how in the world is $B(-1/2, 1) = -2$? I computed it on WolframAlpha, but I am stumped.

$$u = \gamma + u$$

$$0 = \gamma$$

Some error happened unfortunately.

Actually $I \neq \psi(1/2)$. Remember that we have

$$\psi(1/2) = -\gamma -\sum \frac{1}{n(2n-1)}$$

$$I = -\int^1_0 \frac{\log(1-x^2)}{x^2}\,dx=-\sum \frac{1}{n(2n-1)}= u + \gamma$$

, hence $u+\gamma = u +\gamma$ which holds true. As I said you run into a loop by doing this method.

As how to evaluate $B(-1/2,1)$ by definition we have

$$B(-1/2,1) = \frac{\Gamma(-1/2)\Gamma(1)}{\Gamma(1/2)}$$

It is known that $\Gamma(z+1) = z\Gamma(z) $ so $\Gamma(1/2) = (-1/2)\Gamma(-1/2)$.

 

[Amad27]

You will run into a cycle. We want to find the integral without using the digamma function. Fortunately , there is an anti-derivative for the integral. Please , give it more thoughts.
Actually $I \neq \psi(1/2)$. Remember that we have

$$\psi(1/2) = -\gamma -\sum \frac{1}{n(2n-1)}$$

$$I = -\int^1_0 \frac{\log(1-x^2)}{x^2}\,dx=-\sum \frac{1}{n(2n-1)}= u + \gamma$$

, hence $u+\gamma = u +\gamma$ which holds true. As I said you run into a loop by doing this method.

As how to evaluate $B(-1/2,1)$ by definition we have

$$B(-1/2,1) = \frac{\Gamma(-1/2)\Gamma(1)}{\Gamma(1/2)}$$

It is known that $\Gamma(z+1) = z\Gamma(z) $ so $\Gamma(1/2) = (-1/2)\Gamma(-1/2)$.

Okay.

Do you mind giving a small hint?

So will the integral require gamma or beta?

So I can start it knowing a fact =)

Thanks!

 

[alyafey22]

Okay.

Do you mind giving a small hint?

So will the integral require gamma or beta?

So I can start it knowing a fact =)

Thanks!

It can be solved using elemetnary functions. Moreover , you can find an antiderivative of

$$\int \frac{\log(1-x^2)}{x^2}\,dx$$

Hint: start by $t=1/x^2$.

 

[Amad27]

It can be solved using elemetnary functions. Moreover , you can find an antiderivative of

$$\int \frac{\log(1-x^2)}{x^2}\,dx$$

Hint: start by $t=1/x^2$.

How did you find out that specific substition? It is not obvious to start with $t = 1/x^2$ so how did you figure out that we should make that substitution?

It may seem that I am being pedantic, but knowing the reason behind every step is important.

Sorry if this is too unusual, tell me so. =)

 

[alyafey22]

How did you find out that specific substition? It is not obvious to start with $t = 1/x^2$ so how did you figure out that we should make that substitution?

It may seem that I am being pedantic, but knowing the reason behind every step is important.

Sorry if this is too unusual, tell me so. =)

I was going to sleep then I remembered that I made a type. The substitution was supposed to be $t=1/x$. I am really sorry .

As for the reason for the sub , we have $1/x^2$ which is the derivative multiplied by a constant.

 

[alyafey22]

Ok , here is the how to do it

$$\int \frac{\log(1-x^2)}{x^2}\,dx = -\int \log(1-1/t^2) = -\int\log(t^2-1)-2\log(t)\,dt$$

This can be written as

$$-\int\log(t-1)\,dt -\int \log(t+1)\,dt+2\int\log(t)\,dt $$

All the functions above has an ant-derivative.

 

[Amad27]

Ok , here is the how to do it

$$\int \frac{\log(1-x^2)}{x^2}\,dx = -\int \log(1-1/t^2) = -\int\log(t^2-1)-2\log(t)\,dt$$

This can be written as

$$-\int\log(t-1)\,dt -\int \log(t+1)\,dt+2\int\log(t)\,dt $$

All the functions above has an ant-derivative.

Hello,

Okay let's get

$$-\int \log(t-1)\,dt = -(t-1)\ln(t-1) + (t-1)$$
$$-\int \ln(1 + t) dt = -(1+t)\ln(t+1) + (t+1)$$
$$2\int \ln(t) dt = 2t\ln(t) - 2t $$

$$ \implies -(t-1)\ln(t-1) + (t-1) -(1+t)\ln(t+1) + (t+1) + 2t\ln(t) - 2t$$

How can we change the lower bound in terms of $t$

$$ t = 1/x$$
$$t = 1/0$$

Thanks!

 

[alyafey22]

How can we change the lower bound in terms of $t$

$$ t = 1/x$$
$$t = 1/0$$

Thanks!

We just put $t=1/x$. Try then to find the integral on the interval [0,1].

 

[Amad27]

We just put $t=1/x$. Try then to find the integral on the interval [0,1].

But we CANNOT change the lower bound.

$t = 1/0$ because $x = 0$ was the lower bound.

 

[alyafey22]

But we CANNOT change the lower bound.

$t = 1/0$ because $x = 0$ was the lower bound.

The lower and upper limits before the substitution were $0$ and $1$ consecutively. If we make the substitution they are $\infty$ and $1$. You have the choice between $x$ and $t$. Remember if we have an improper integral we take the limit

$$\int^\infty_1 f(x) \,dx = \lim_{t \to \infty }\int^t_1 f(x) \,dx$$

 

[Amad27]

The lower and upper limits before the substitution were $0$ and $1$ consecutively. If we make the substitution they are $\infty$ and $1$. You have the choice between $x$ and $t$. Remember if we have an improper integral we take the limit

$$\int^\infty_1 f(x) \,dx = \lim_{t \to \infty }\int^t_1 f(x) \,dx$$

Hello,

@ZaidAlyafey, I hope you did not misunderstand that I was trying to ignore you. I did indeed read this reply, but I stood off for a while, getting more in depth with this topic, so now I can discuss this. Please do not mind, I appreciate every bit and I won't give up on this, until I figure it out! =)

I have labeled sections of questions, I would appreciate if you respond to each section separately.

QUESTION #1:
You had this: (We were trying to evaluate $\psi(1)$ as an example)

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$

How did you arrive at this result?

QUESTION #2:
Secondly: You stated the following:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

But you did not integrate the LHS. Can you take me behind the scenes here?

QUESTION #3:
Please tell me. Can you write sums as integrals for every sum? If so, how? Thanks!

Final comment:
Again, you did not give up on me, and I did not give up on you either! I just took a break so I could read more (as you have seen from my posts). Please do not mind this, I try not to give up on anything! Thanks for your dedication.

 

[alyafey22]

QUESTION #1:
You had this: (We were trying to evaluate $\psi(1)$ as an example)

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$

How did you arrive at this result?

Ok we start by the following

$$\sum_{n=1}^\infty x^{n-1} = \frac{1}{1-x}$$

Then integrate to get

$$\sum_{n=1}^\infty \frac{x^n}{n} =- \log(1-x)$$

Finally let $x= z^2$

---

QUESTION #2:
Secondly: You stated the following:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

But you did not integrate the LHS. Can you take me behind the scenes here?

From the first part we proved that

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Now divide by $z^2$ to obtain

$$\sum_{n=1}^\infty \frac{z^{2n-2}}{n} = -\frac{\log(1-z^2)}{z^2}$$

Now integrate with respect to $z$

$$\sum_{n=1}^\infty \int^z_0 \frac{z^{2n-2}}{n} \,dz= -\int^z_0\frac{\log(1-z^2)}{z^2}\,dz$$

---

QUESTION #3:
Please tell me. Can you write sums as integrals for every sum? If so, how? Thanks!

Definite integrals are Riemann sums , so it must be possible but might be complicated. Moreover , it will not necessarily make the problem easier to solve.

---

Final comment:
Again, you did not give up on me, and I did not give up on you either! I just took a break so I could read more (as you have seen from my posts). Please do not mind this, I try not to give up on anything! Thanks for your dedication.

It is good that you took your time trying to understand series.

 

[Amad27]

Ok we start by the following

$$\sum_{n=1}^\infty x^{n-1} = \frac{1}{1-x}$$

Then integrate to get

$$\sum_{n=1}^\infty \frac{x^n}{n} =- \log(1-x)$$

Finally let $x= z^2$

---

From the first part we proved that

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Now divide by $z^2$ to obtain

$$\sum_{n=1}^\infty \frac{z^{2n-2}}{n} = -\frac{\log(1-z^2)}{z^2}$$

Now integrate with respect to $z$

$$\sum_{n=1}^\infty \int^z_0 \frac{z^{2n-2}}{n} \,dz= -\int^z_0\frac{\log(1-z^2)}{z^2}\,dz$$

---

Definite integrals are Riemann sums , so it must be possible but might be complicated. Moreover , it will not necessarily make the problem easier to solve.

---

It is good that you took your time trying to understand series.

It has been known that:

QUESTION #1:

$\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$

Let $u = n-1$ so

$\displaystyle \sum_{n=1}^{\infty} x^{n-1} = \sum_{u=0}^{\infty} x^{u} = \frac{1}{1-x}$

Is this how you derive that result?

QUESTION #2:
About interchanging summation and integral, doesn't the series have to be a monotone series? How did you test if that series was monotone?
QUESTION #3:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

You cannot simply get $\psi(1)$ just be evaluating the integral. The LHS is NOT the proper definition of the digamma $\psi$ function. Wikipedia states:

$\psi(z) = \gamma + \displaystyle \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$

So what next?

Thanks Zaid!

 

[alyafey22]

It has been known that:

QUESTION #1:

$\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$

Let $u = n-1$ so

$\displaystyle \sum_{n=1}^{\infty} x^{n-1} = \sum_{u=0}^{\infty} x^{u} = \frac{1}{1-x}$

Is this how you derive that result?

Yes , exactly.

QUESTION #2:

About interchanging summation and integral, doesn't the series have to be a monotone series? How did you test if that series was monotone?

The function is a geometric series , so you can integrate term by term.

QUESTION #3:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

You cannot simply get $\psi(1)$ just be evaluating the integral. The LHS is NOT the proper definition of the digamma $\psi$ function. Wikipedia states:

$\psi(z) = \gamma + \displaystyle \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$

Revise post #23.

 

[Amad27]

Yes , exactly.
The function is a geometric series , so you can integrate term by term.
Revise post #23.

Hello, I just reread that post (long time ago!)

So let $z = 1$ then you have the required sum? That makes sense.

I suppose you have tons of practice with this stuff. I just can't think like that. Perhaps its the lack of my practice (maybe due to my age). Thanks though. Let's try another one?

$\psi(4) = \psi(3 + 1)$, so $z = 3$

$\psi(z+1) = \gamma + \displaystyle \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$

When $z = 3$, $\psi(3+1) = \gamma + \displaystyle 3\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$

$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

$\displaystyle \sum_{u=2}^{\infty} x^{u} = \frac{1}{1-x} - (x + x^2)$

Where $u = n + 2$

$\displaystyle \sum_{u=2}^{\infty} \int_{0}^{z-3} x^{u} \,dx = \int_{0}^{z-3} \frac{1}{1-x} - (x + x^2) \,dx$

$= \displaystyle \sum_{u=2}^{\infty} \frac{(z-3)^{u+1}}{u+1} = \int_{0}^{z-3} \frac{1}{1-x} - (x + x^2) \,dx$

$= \displaystyle \sum_{n=0}^{\infty} \frac{(z-3)^{n+3}}{n+3} = \int_{0}^{z-3} \frac{1}{1-x} - (x + x^2) \,dx$

$= \displaystyle \sum_{n=0}^{\infty} \frac{(z-3)^{n+3}}{z(n+3)} = \displaystyle \frac{\int_{0}^{z-3} \frac{1}{1-x} - (x + x^2) \,dx}{z}$
Anything that may help?

Thanks!

 

[alyafey22]

We have already proved that

$$\sum_{n=1}^\infty \frac{z^n }{n} = -\log(1-z)$$

Multiply by $z^2$ to get

$$\sum_{n=1}^\infty \frac{z^{n+2} }{n} = -z^2\log(1-z)$$

Now integrate

$$\sum_{n=1}^\infty \frac{z^{n+3} }{n(n+3)} = -\int^z_0 x^2\log(1-x)\,dx$$

Now set $z=1$

$$\sum_{n=1}^\infty \frac{1 }{n(n+3)} = -\int^1_0 x^2\log(1-x)\,dx$$

That should be easy solvable using $t = 1-x$.

Note you can also use

$$\psi(z+1) = \psi(z)+\frac{1}{z}$$.

 

[Amad27]

We have already proved that

$$\sum_{n=1}^\infty \frac{z^n }{n} = -\log(1-z)$$

Multiply by $z^2$ to get

$$\sum_{n=1}^\infty \frac{z^{n+2} }{n} = -z^2\log(1-z)$$

Now integrate

$$\sum_{n=1}^\infty \frac{z^{n+2} }{n(n+3)} = -\int^z_0 x^2\log(1-x)\,dx$$

Now set $z=1$

$$\sum_{n=1}^\infty \frac{1 }{n(n+3)} = -\int^1_0 x^2\log(1-x)\,dx$$

That should be easy solvable using $t = 1-x$.

Note you can also use

$$\psi(z+1) = \psi(z)+\frac{1}{z}$$.

That is fantastic. Let's find

$\psi(5/4) = \psi(1 + 1/4)$ therefore, $z = 1/4$

There is one issue with the integration (not yours but in general).

When you integrate you change the $Z$'s in the integrand to $x$'s. But you do not do the same on the RHS. When you do:

$$\sum_{n=1}^\infty \frac{z^{n+2} }{n} = -z^2\log(1-z)$$

Now integrate

$$\sum_{n=1}^\infty \frac{z^{n+2} }{n(n+3)} = -\int^z_0 x^2\log(1-x)\,dx$$

How?

 

[alyafey22]

That is fantastic. Let's find

$\psi(5/4) = \psi(1 + 1/4)$ therefore, $z = 1/4$

Let us see what you have tried.

When you integrate you change the $Z$'s in the integrand to $x$'s. But you do not do the same on the RHS. When you do:

How?

I made a mistake it should be

$$\sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} =- \int^z_0 x^2\log(1-x)\,dx$$

Since it is a dummy variable you can use $z$ but I am trying to avoid confusion. The following is also correct

$$\sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} = -\int^z_0 z^2\log(1-z)\,dz$$

 

[Amad27]

Let us see what you have tried.
I made a mistake it should be

$$\sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} =- \int^z_0 x^2\log(1-x)\,dx$$

Since it is a dummy variable you can use $z$ but I am trying to avoid confusion. The following is also correct

$$\sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} = -\int^z_0 z^2\log(1-z)\,dz$$

So then,

Let $z = 1$ On the RHS you have,

$$\sum_{n=1}^\infty\frac{1}{n(n+3)} = -\int^1_0 \log(1-1)\,dz$$

which is not possible.

 

[alyafey22]

So then,

Let $z = 1$ On the RHS you have,

$$\sum_{n=1}^\infty\frac{1}{n(n+3)} = -\int^1_0 \log(1-1)\,dz$$

which is not possible.

This is where a confusion happens. The inner $z$ is independent of the outer $z$ so to avoid that we always avoid such symbols. The inner $z$ is a dummy variable so we can choose any thing we want $x$ , $y$ or even $z$. So it is always preferable to use a symbol different than the boundary of integration for example $x$

$$\int^z_0 x^2\log(1-x)\,dx$$

 

[Amad27]

This is where a confusion happens. The inner $z$ is independent of the outer $z$ so to avoid that we always avoid such symbols. The inner $z$ is a dummy variable so we can choose any thing we want $x$ , $y$ or even $z$. So it is always preferable to use a symbol different than the boundary of integration for example $x$

$$\int^z_0 x^2\log(1-x)\,dx$$

Why not just avoid all confusion, and directly integrate form 0 to 1?anyway,

$\psi(5/4), z = 1/4$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)$

Let $x = z^4$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(4n+1)} = -\int_{0}^{1} \ln(1-z^4) \,dz$

So we must figure out $\displaystyle -\int_{0}^{1} \ln(1-z^4) \,dz$ then add $\gamma$ to it. The integral is tough though.

The antiderivative is too intense to even look at. How would you evaluate the integral?

Out of curiosity, can you get a closed-form by using series. Maybe that could help. I was looking at the challenges section of the forum.

One of them is your logarithmic integral #3, which remains unsolved..

I just had a thought... if you don't mind, maybe we can try the integral, since no one has done it in two months or is that not allowed? Thanks.

 

[alyafey22]

Why not just avoid all confusion, and directly integrate form 0 to 1?

Sure , that might be better :)

anyway,

$\psi(5/4), z = 1/4$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)$

Let $x = z^4$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(4n+1)} = -\int_{0}^{1} \ln(1-z^4) \,dz$

So we must figure out $\displaystyle -\int_{0}^{1} \ln(1-z^4) \,dz$ then add $\gamma$ to it. The integral is tough though.

The antiderivative is too intense to even look at. How would you evaluate the integral?

It might not be as tought as you think , use the property

$$(1-z^4) = (1-z)(1+z)(1+z^2)$$

Out of curiosity, can you get a closed-form by using series. Maybe that could help. I was looking at the challenges section of the forum.

One of them is your logarithmic integral #3, which remains unsolved..

I just had a tough... if you don't mind, maybe we can try the integral, since no one has done it in two months or is that not allowed? Thanks.

Can you give me the link ?

 

[Amad27]

Sure , that might be better :)
It might not be as tought as you think , use the property

$$(1-z^4) = (1-z)(1+z)(1+z^2)$$
Can you give me the link ?

Hey there!

$\displaystyle \log((1-z)(1+z)(1+z^2)) = \log(1-z) + \log(1+z) + \log(1+z^2)$

$\displaystyle \int_{0}^{1} \log((1-z)(1+z)(1+z^2)) \,dz = \int_{0}^{1} \log(1-z) + \log(1+z) + \log(1+z^2) \,dz = \lim_{z\to 1} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + \int_{0}^{1} \log(1 + z^2) \,dz$

Currently, I am running out of sleep, so I'll continue the integration tomorrow. Please do not give hints, just let me know if the setup is correct. Thanks.

NEXT:
----------
The integral was: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html

I spent hours and hours trying this, but later find out it requires special functions...

Question #1:
Do you know where I can get a quick course on special functions and more IMPORTANTLY, how to implement them into integration?

Functions such as $\zeta(s), L_1, L_2$ I am not sure what the ones with the $L$ are, supposed logarithmic integrals for $\frac{1}{\ln(x)}$ But perhaps you could guide me to somewhere they talk about this and how to implement this in depth.

Question #2:
You posted a hint there as well. I assume you want me to integrate the series, LHS, RHS to get it into some sort of form?

Question #3:
What is that series anyway? I am not familiar with the $H_kx^k$ idea or what It is. I tried putting it online, but no result?

Question #4:
Can you perhaps show me a page where they talk about all these laws, such as the one with dilogs etc shown here: http://mathhelpboards.com/calculus-10/higher-order-reflected-logarithms-13087.html

Thank you @ZaidAlyafey, I appreciate your guidance here in this forum, it is highly valuable to me! Thanks a bunch =)

 

[alyafey22]

Hey there!

$\displaystyle \log((1-z)(1+z)(1+z^2)) = \log(1-z) + \log(1+z) + \log(1+z^2)$

$\displaystyle \int_{0}^{1} \log((1-z)(1+z)(1+z^2)) \,dz = \int_{0}^{1} \log(1-z) + \log(1+z) + \log(1+z^2) \,dz = \lim_{z\to 1} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + \int_{0}^{1} \log(1 + z^2) \,dz$

Currently, I am running out of sleep, so I'll continue the integration tomorrow. Please do not give hints, just let me know if the setup is correct. Thanks.

Let me see the full solution.

NEXT:
----------
The integral was: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html

I spent hours and hours trying this, but later find out it requires special functions...

There are many ways to solve that integral. You can solve it using the beta function. It is a good practice , try it .

Question #1:
Do you know where I can get a quick course on special functions and more IMPORTANTLY, how to implement them into integration?

Functions such as $\zeta(s), L_1, L_2$ I am not sure what the ones with the $L$ are, supposed logarithmic integrals for $\frac{1}{\ln(x)}$ But perhaps you could guide me to somewhere they talk about this and how to implement this in depth.

These are called dilogarithms. I discussed lots of special functions on this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post14488.html#post14488.

Question #2:
You posted a hint there as well. I assume you want me to integrate the series, LHS, RHS to get it into some sort of form?

Exactly.

Question #3:
What is that series anyway? I am not familiar with the $H_kx^k$ idea or what It is. I tried putting it online, but no result?

$H_k$ is called the harmonic number. The series containg them are called Euler sums.

Question #4:
Can you perhaps show me a page where they talk about all these laws, such as the one with dilogs etc shown here: http://mathhelpboards.com/calculus-10/higher-order-reflected-logarithms-13087.html

You can refer to the link above on Advanced integration techniques. In this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post23071.html#post23071I discuss polylogarithms.

 

[Amad27]

Let me see the full solution.
There are many ways to solve that integral. You can solve it using the beta function. It is a good practice , try it .
These are called dilogarithms. I discussed lots of special functions on this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post14488.html#post14488.
Exactly.
$H_k$ is called the harmonic number. The series containg them are called Euler sums.
You can refer to the link above on Advanced integration techniques. In this http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233-post23071.html#post23071I discuss polylogarithms.

The improper integrals must be divergent. The limit cannot exist because of $\ln(1-z)$ the limit as $x \to 1$ does not exist.

Question #1:
Also where did you learn special functions? What is a possible textbook, since i can access that offline. Your page was amazing but my Internet can't load all LaTex making it immensely hard to read. But I tried.

I read the page, I was able to prove the integral.
$\displaystyle Li_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2} = -\int_{0}^{z} \frac{log(1-x)}{x}\,dx$

$\displaystyle \sum_{n=1}^{\infty} {x}^{n-1} = \frac{1}{1-x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln\left({1-x}\right)$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = \frac{-\ln(1-x)}{x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{z^n}{n^2} = \int_{0}^{z} \frac{-\ln(1-x)}{x} \,dx$

Let me know.

Question #2:

How can you integrate $Li_2$? I tried from $0 \to 1$

$\displaystyle \int_{0}^{1} Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$

$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$

$$(An + B)(n+1) + D(n^2) = 1$$

Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$

$$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$

$$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$

The $1/n$ is the problem, it is the harmonic series, which diverges.

Thanks (BTW: Where did you find the dilog reflection law?)

Thanks =)

 

[alyafey22]

The improper integrals must be divergent. The limit cannot exist because of $\ln(1-z)$ the limit as $x \to 1$ does not exist.

I must see your full solution. If you are talking about $\int^1_0\log(1-z)dz$ then this integral converges and it is value is $-1$.

Question #1:
Also where did you learn special functions? What is a possible textbook, since i can access that offline. Your page was amazing but my Internet can't load all LaTex making it immensely hard to read. But I tried.

Oh , that is so bad. Unfortunately , I learned each special function separately by searching the internet about its properties. I don't reckon seeing a book that explains all special functions.

I read the page, I was able to prove the integral.
$\displaystyle Li_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2} = -\int_{0}^{z} \frac{log(1-x)}{x}\,dx$

$\displaystyle \sum_{n=1}^{\infty} {x}^{n-1} = \frac{1}{1-x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln\left({1-x}\right)$

$\displaystyle \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = \frac{-\ln(1-x)}{x}$

$\displaystyle \sum_{n=1}^{\infty} \frac{z^n}{n^2} = \int_{0}^{z} \frac{-\ln(1-x)}{x} \,dx$

Let me know.

That is good , you are making a big progress.

Question #2:

How can you integrate $Li_2$? I tried from $0 \to 1$

$\displaystyle \int_{0}^{1} Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$

$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$

$$(An + B)(n+1) + D(n^2) = 1$$

Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$

$$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$

$$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$

The $1/n$ is the problem, it is the harmonic series, which diverges.

Very nice. Now separate the sum into

$$\sum \frac{1}{n^2 }+\sum \frac{1}{n+1}-\frac{1}{n}$$

The right is a telescoping series , isn't it ?

Thanks (BTW: Where did you find the dilog reflection law?)

Thanks =)

http://mathhelpboards.com/challenge-questions-puzzles-28/dilogarithmic-integration-parts-10876.html?highlight=reflectionare two proofs.

 

[Amad27]

I must see your full solution. If you are talking about $\int^1_0\log(1-z)dz$ then this integral converges and it is value is $-1$.
Oh , that is so bad. Unfortunately , I learned each special function separately by searching the internet about its properties. I don't reckon seeing a book that explains all special functions.
That is good , you are making a big progress.
Very nice. Now separate the sum into

$$\sum \frac{1}{n^2 }+\sum \frac{1}{n+1}-\frac{1}{n}$$

The right is a telescoping series , isn't it ?
http://mathhelpboards.com/challenge-questions-puzzles-28/dilogarithmic-integration-parts-10876.html?highlight=reflectionare two proofs.

Thank you ZaidAlyafey, you are a very encouraging person, which I appreciate. I appreciate your help here as well, I have learned a lot from you, especially about series.

I went to the proof page. I read your proof, it was very nice :D But it stated,

take $x = 1$ we get C = $Li_2$, but how? What is $f(x)$ if $x = 1$? I won't be defined.

The integral first.

I need $\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz$ But the question is how...

$\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz = ?$

I am thinking of a strategy for this. I checked WolframAlpha. The integral is quite intense, I think by parts is the best idea.

It was intense, I did it on paper. I ran into a loop, typing it would be very hard, I'll do the last few steps here.

$I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx) + 0.5I]$

$-I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

$I = \displaystyle -x\ln(1 + x^2) + 2x^2\arctan(x) - 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

Finally,

$I_{Final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [(-1\ln(2)) + 2\arctan(1) - 4[\arctan(1) - 0.5\log(2) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - 4[\pi/4 - \log(\sqrt{2}) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - \pi + \log(4) - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [\log(2) - \frac{\pi}{2} - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$$2\int x\arctan(x) \,dx = x^2\arctan(x) - (x/2)\log(1+x^2) + 0.5\int log(1 + x^2)$$

Ahhh this is crazy. So much going around, now I am back at the log(1+x^2) integral... Can this be done otherwise?

Lets try the: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html integral.

$\displaystyle \int_{0}^{1} \log^2(x)\log^2(1-x) \,dx$

So $\displaystyle \sum_{k=1}^{\infty} H_kx^k = \frac{-\log(1-x)}{1-x}$

$$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

$$\sum_{n=1}^{\infty} -\log(1-x)x^{n-1} = \frac{-\log(1-x)}{1-x}$$

If it has to do with integration of $H_k$ then I have no clue. Well, I sort of do.

$\displaystyle H_k = \sum_{n=1}^{k} \frac{1}{n}$

$\displaystyle (H_k)(x^k) = \sum_{n=1}^{k} \frac{x^k}{n}$

$\displaystyle \sum_{k=1}^{n} H_nx^n = \frac{-\log(1-x)}{1-x}$

$\displaystyle \sum_{k=1}^{\infty} \int_{0}^{z}H_k(x^k) \,dx = \int_{0}^{z} \frac{-\log(1-x)}{1-x} \,dx$

I have just one issue.

$H_k = \displaystyle \sum_{n=1}^{k} \frac{1}{n}$
$x^k = x^k$

$H_k (x^k) = \displaystyle \sum_{n=1}^{k} \frac{x^k}{n)$

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n}$ is what we must figure out, which seems ultimately difficult.

using Fubini's and Tonelli's theorems,

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n} = \sum_{n=1}^{k} \frac{1}{n} \sum_{k=1}^{\infty} \frac{x^k}{1}$

$\displaystyle \sum_{k=1}^{\infty} x^{k} = \frac{1}{1-x} - 1 = \frac{x}{1-x}$

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n} = \sum_{n=1}^{k} \frac{1}{n} \sum_{k=1}^{\infty} \frac{x^k}{1} = \sum_{n=1}^{k} \frac{x}{(1-x)(n)} = \frac{xH_k}{1-x}$

Okay, I am confused.

How can you integrate $H_k$? It becomes extremely difficult. Any suggestions? Thanks

 

[alyafey22]

I went to the proof page. I read your proof, it was very nice :D But it stated,

take $x = 1$ we get C = $Li_2$, but how? What is $f(x)$ if $x = 1$? I won't be defined.

I started by setting

$$f(x) = \mathrm{Li}_2(x) + \mathrm{Li}_2(1-x)$$

Hence setting $x\to 1 $ we get

$$c = \lim_{x\to 1 } f(x) +\log(x)\log(1-x) = \mathrm{Li}_2(1) + \mathrm{Li}_2(0)+ 0 = \mathrm{Li}_2(1) =\sum \frac{1}{n^2} = \zeta(2)$$

The integral first.

I need $\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz$ But the question is how...

$\displaystyle \int_{0}^{1} \ln(1+ z^2) \,dz = ?$

I am thinking of a strategy for this. I checked WolframAlpha. The integral is quite intense, I think by parts is the best idea.

It was intense, I did it on paper. I ran into a loop, typing it would be very hard, I'll do the last few steps here.

$I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx) + 0.5I]$

$-I = \displaystyle x\ln(1 + x^2) - 2x^2\arctan(x) + 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

$I = \displaystyle -x\ln(1 + x^2) + 2x^2\arctan(x) - 4[x(x\arctan(x) - 0.5\log(1+x^2)) - \int (x\arctan(x) \space dx)]$

Finally,

$I_{Final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [(-1\ln(2)) + 2\arctan(1) - 4[\arctan(1) - 0.5\log(2) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - 4[\pi/4 - \log(\sqrt{2}) - \int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [-\log(2) + \pi/2 - \pi + \log(4) - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$I_{final} = \displaystyle \lim_{{z}\to{1}} -(1-z)(\log(1-z) - 1) + (1+z)(\log(1+z) - 1) + [\log(2) - \frac{\pi}{2} - 4\int_{0}^{1} x\arctan(x) \,dx]]$

$$2\int x\arctan(x) \,dx = x^2\arctan(x) - (x/2)\log(1+x^2) + 0.5\int log(1 + x^2)$$

Ahhh this is crazy. So much going around, now I am back at the log(1+x^2) integral... Can this be done otherwise?

Maybe you are doing it the wrong way ? Integrate $dx$ and differentiate $\log(1+x^2)$

Lets try the: http://mathhelpboards.com/challenge-questions-puzzles-28/not-solved-logarithm-integral-3-a-12050.html integral.

$\displaystyle \int_{0}^{1} \log^2(x)\log^2(1-x) \,dx$

So $\displaystyle \sum_{k=1}^{\infty} H_kx^k = \frac{-\log(1-x)}{1-x}$

$$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

$$\sum_{n=1}^{\infty} -\log(1-x)x^{n-1} = \frac{-\log(1-x)}{1-x}$$

If it has to do with integration of $H_k$ then I have no clue. Well, I sort of do.

$\displaystyle H_k = \sum_{n=1}^{k} \frac{1}{n}$

$\displaystyle (H_k)(x^k) = \sum_{n=1}^{k} \frac{x^k}{n}$

$\displaystyle \sum_{k=1}^{n} H_nx^n = \frac{-\log(1-x)}{1-x}$

$\displaystyle \sum_{k=1}^{\infty} \int_{0}^{z}H_k(x^k) \,dx = \int_{0}^{z} \frac{-\log(1-x)}{1-x} \,dx$

I have just one issue.

$H_k = \displaystyle \sum_{n=1}^{k} \frac{1}{n}$
$x^k = x^k$

$H_k (x^k) = \displaystyle \sum_{n=1}^{k} \frac{x^k}{n)$

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n}$ is what we must figure out, which seems ultimately difficult.

using Fubini's and Tonelli's theorems,

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n} = \sum_{n=1}^{k} \frac{1}{n} \sum_{k=1}^{\infty} \frac{x^k}{1}$

$\displaystyle \sum_{k=1}^{\infty} x^{k} = \frac{1}{1-x} - 1 = \frac{x}{1-x}$

$\displaystyle \sum_{k=1}^{\infty} \displaystyle \sum_{n=1}^{k} \frac{x^k}{n} = \sum_{n=1}^{k} \frac{1}{n} \sum_{k=1}^{\infty} \frac{x^k}{1} = \sum_{n=1}^{k} \frac{x}{(1-x)(n)} = \frac{xH_k}{1-x}$

Okay, I am confused.

How can you integrate $H_k$? It becomes extremely difficult. Any suggestions? Thanks

I am confused too :) . It will be difficult for you trying to solve many different problems using different approaches. Euler sums require a descent knowledge of polylogarithms. Let us stick with finding the integral by differentiating Beta ?

Maybe we can start out by and easier one

$$\int^1_0 \log(x)\log(1-x) \, dx$$

PS: Nice that you are trying to post your attempts

 

[Amad27]

I started by setting

$$f(x) = \mathrm{Li}_2(x) + \mathrm{Li}_2(1-x)$$

Hence setting $x\to 1 $ we get

$$c = \lim_{x\to 1 } f(x) +\log(x)\log(1-x) = \mathrm{Li}_2(1) + \mathrm{Li}_2(0)+ 0 = \mathrm{Li}_2(1) =\sum \frac{1}{n^2} = \zeta(2)$$
Maybe you are doing it the wrong way ? Integrate $dx$ and differentiate $\log(1+x^2)$

I am confused too :) . It will be difficult for you trying to solve many different problems using different approaches. Euler sums require a descent knowledge of polylogarithms. Let us stick with finding the integral by differentiating Beta ?

Maybe we can start out by and easier one

$$\int^1_0 \log(x)\log(1-x) \, dx$$

PS: Nice that you are trying to post your attempts

Thank you Zaid. I will post my attempts on the first integral (the one with $\ln(1-z^4)$) tomorrow, I want to try this second one now.

$$\int^1_0 \log(x)\log(1-x) \, dx$$

I actually asked $Li_2$ because of this, we notice that.

$\log(x)\log(1-x) = -Li_2(x) - Li_2(1-x) + \pi^2/6$

$$\int^1_0 \log(x)\log(1-x) \, dx = \int_{0}^{1} -Li_2(x) - Li_2(1-x) + \pi^2/6 \,d$$

$Li_2(x) = \displaystyle \sum_{k=1}^{\infty} \frac{x^k}{k^2}$

$\displaystyle \int_{0}^{1} Li_2(x) \,dx = \displaystyle \sum_{k=1}^{\infty} \int_{0}^{1} \frac{x^k}{k^2} \,dx = \sum_{k=1}^{\infty} \frac{1}{k^2(k+1)} $

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2(k+1)} = \sum_{k=1}^{\infty}\frac{1}{k^2} - \frac{1}{k} + \frac{1}{(k+1)}$

$\displaystyle = \zeta(2) + \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k}$

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k}$

$S_1 = \frac{1}{2} - 1 $
$S_2 = \frac{1}{2} - 1 + \frac{1}{3} - \frac{1}{2}$
$S_3 = \frac{1}{2} - 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{4} - \frac{1}{3}$

$S_n = -1 + \frac{1}{n+1}$
$\displaystyle \lim_{n\to\infty} S_n = -1$

$Sum_{final} = \zeta(2) - 1$

Next We go to $\Li_2(1-x)$

$\Li_2(1-x) = \displaystyle \sum_{k=1}^{\infty} \frac{(1-x)^k}{k^2}$

$\int_{0}^{1} \Li_2(1-x) \,dx = \displaystyle \sum_{k=1}^{\infty}\int_{0}^{1} \frac{(1-x)^k}{k^2} \,dx$

$$ = \sum_{k=1}^{\infty} 0 = 0$$

because $Li_2(x)$ works for $0 < x < 1$,
$Li_2(1-x)$ should have domain $ 1 > x > 0$

So we take the limit as $x \to 1$ which is $-infty$ This is the hard part.

If I could get some help with $Li_2(1-x)$ I should be ready to roll!

Thanks. I would like to try the integral with Euler sums rather than differentiating Beta. Mostly because the Euler method is cool. And second, I have never learned multivariable calculus formally, just from some web reading, so when it comes to differentials (multivariable, will be required) I won't be able to do anything. So I would like to try euler sums!

 

[alyafey22]

Use \mathrm{Li}_p to display polylogarithms in a better way.

 

[Amad27]

Could you rewrite that? It doesn't display well on my machine.

Its okay. I got it,

$\displaystyle \lim_{x\to 1} (1-x) = 0$

$Li_2(1-x) = Li_2(x) = \zeta(2) - 1$

$\displaystyle = \int_{0}^{1} -(\zeta(2) - 1) + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} -2\zeta(2) + 2 + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} 2 - \zeta(2) \,dx$

$= 2 - \zeta(2)$

Can we try the second power integral with Euler Sums ?? Please?

 

[alyafey22]

Its okay. I got it,

$\displaystyle \lim_{x\to 1} (1-x) = 0$

$Li_2(1-x) = Li_2(x) = \zeta(2) - 1$

$\displaystyle = \int_{0}^{1} -(\zeta(2) - 1) + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} -2\zeta(2) + 2 + \zeta(2) \,dx$

$\displaystyle = \int_{0}^{1} 2 - \zeta(2) \,dx$

$= 2 - \zeta(2)$

Some how you got the idea but it should be like this

$$\int^1_0 \mathrm{Li}_2(1-t) \,dt = \int^1_0 \mathrm{Li}_2(x)\,dx $$

using the change of the variable $t = 1-x$.

Can we try the second power integral with Euler Sums ?? Please?

Ok, but you have to revise your calculations and focus more. Let me see your second attempt.

 

[Amad27]

Some how you got the idea but it should be like this

$$\int^1_0 \mathrm{Li}_2(1-t) \,dt = \int^1_0 \mathrm{Li}_2(x)\,dx $$

using the change of the variable $t = 1-x$.
Ok, but you have to revise your calculations and focus more. Let me see your second attempt.

Okay, I'll do a full version here:

$$\int_{0}^{1} \log(x)\log(1-x) \,dx$$
$$\log(x)\log(1-x) = -(L_2(x) + L_2(1-x)) + \zeta(2)$$
$$= \int_{0}^{1} -(L_2(x) + L_2(1-x)) \,dx + \zeta(2)$$

Evaluation of $\int_{0}^{1} L_2(x) \,dx$:
$$L_2(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}$$
$$\int_{0}^{1} L_2(x) \,dx = \sum_{k=1}^{\infty} \frac{1}{k^2(k+1)}$$
$$= \zeta(2) + \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k^2}$$
$$S_1 = \frac{1}{2} - \frac{1}{1}$$
$$S_n = \frac{1}{n+1} - 1$$
$$\lim_{n\to\infty} S_n = -1$
Integral of $L_2(x) = \zeta(2) - 1$

Evaluation of $L_2(1-x)$:

The sum is exactly like the one as $L_2(x)$ so integral of $L_2(1-x) = \zeta(2) - 1$

Evaluation of all combined:
$$= \int_{0}^{1} -(L_2(x) + L_2(1-x)) \,dx + \zeta(2)$$
$$= \int_{0}^{1} -2(\zeta(2) - 1) \,dx + \zeta(2)$$
$$ = -2\zeta(2) + 2 + \zeta(2)$$
$$I_{final} = 2 - \zeta(2)$$

I began working on the power integral.

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx$$

$$\sum_{k=1}^{\infty} \frac{x^k}{k} = -\frac{\log(1-x)}{1-x}$$
$$\int\sum_{k=1}^{\infty} \frac{x^k}{k} \,dx = \int -\frac{\log(1-x)}{1-x} \,dx$$
Let $u = \log(1-x) \implies du = \frac{-1}{1-x} dx$
$$\int\sum_{k=1}^{\infty} \frac{x^k}{k} \,dx = \frac{\log^2(1-x)}{2} $$
$$\log^2(1-x) = 2\sum_{k=1}^{\infty} \frac{x^{k+1}}{k(k+1)}$$

We see that:

$$\sum_{k=1}^{\infty} \frac{z^k}{k} = -\frac{\log(1-z)}{1-z}$$
Let $z = 1 - x$, it guarantees $|z| < 1$
$$(-)\sum_{k=1}^{\infty} \frac{(x+1)^k}{k} = \frac{\log(x)}{x}$$
$$(-2)\sum_{k=1}^{\infty} \frac{(x+1)^{k+1}}{k(k+1)} = \log^2(x)$$

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx = (-4)\cdot\int_{0}^{1} \sum_{k=1}^{\infty} \frac{(x+1)^{k+1}}{k(k+1)})(\sum_{k=1}^{\infty} \frac{x^{k+1}}{k(k+1)})\,dx$$

So far so good?

$$= (-4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{(x+1)^{k+1}}{k(k+1)}) \frac{x^{k+1}}{k(k+1)}) \,dx = (-4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} (\frac{(2)^{k+1}}{k(k+1)}) - \frac{(1)}{k(k+1)}))(\frac{1}{k(k+1)}) = (-4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \frac{(2^{k+1} - 1)}{k^2(k+1)^2}$$

I must've done something wrong somewhwre.

Ah-Ha I found the mistake. I got the wrong square log.

It still won't be correct. I just can't get the squared log.

 

[alyafey22]

Evaluation of all combined:
$$= \int_{0}^{1} -(L_2(x) + L_2(1-x)) \,dx + \zeta(2)$$
$$= \int_{0}^{1} -2(\zeta(2) - 1) \,dx + \zeta(2)$$
$$ = -2\zeta(2) + 2 + \zeta(2)$$
$$I_{final} = 2 - \zeta(2)$$

You should not write it like that. It should be

$$ -\int^1_0 L_2(x)dx- \int^1_0 L_2(1-x) dx = -2\int^1_0 L_2(x)dx = -2(\zeta(2)-1) $$

$$\sum_{k=1}^{\infty} \frac{x^k}{k} = -\frac{\log(1-x)}{1-x}$$

How did you get that ? are you missing the harmonic number ?

 

[Amad27]

You should not write it like that. It should be

$$ -\int^1_0 L_2(x)dx- \int^1_0 L_2(1-x) dx = -2\int^1_0 L_2(x)dx = -2(\zeta(2)-1) $$
How did you get that ? are you missing the harmonic number ?

mmm... Why can't I write the integral like that? I just factored out a $(-1)$ Is that not correct?

Secondly.

I did miss out the harmonic number unforunately. But anyhow.

$$\sum_{k=1}^{\infty} (H_k)(x^k) = -\frac{\log(1-x)}{1-x}$$

$$\sum_{k=1}^{\infty}\left(\sum_{n=1}^{k} \frac{1}{n}\right)\left(x^k\right) = \frac{\log(1-x)}{x-1}$$

Actually,

$$\sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} = \frac{-\log^2(1-x)}{2} \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $$

Since $\displaystyle \sum_{k=1}^{\infty} (H_k)(x^k) = -\frac{\log(1-x)}{1-x}$ this implies the fact that $\displaystyle \sum_{k=1}^{\infty} (H_k)((1-x)^k) = -\frac{\log(x)}{x}$

$$ (-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1} = \log^2(x)$$

$$\implies \log^2(x)\log^2(1-x) = \left((-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left((-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right)$$

$$\implies \int_{0}^{1}\log^2(x)\log^2(1-x) \,dx = \int_{0}^{1} \left((-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left((-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot\int_{0}^{1} \left( \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left( \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot \left( \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{(H_k)(1-x)^{k+1}}{k+1} \cdot \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot ( \sum_{k=1}^{\infty} \sum_{k=1}^{\infty})\frac{(H_k)^2}{(k+1)^2} \cdot \int_{0}^{1} (1-x)^{k+1}\cdot(x)^{k+1}\,dx$$

Realize, $\displaystyle \int_{0}^{1} (1-x)^{(k+2) - 1}\cdot(x)^{(k+2) - 1} \,dx = B(k+2, k+2) = \frac{\Gamma^2(k+2)}{\Gamma(2(k+2)}$

Finally, $\displaystyle = (4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty}\frac{(H_k)^2}{(k+1)^2} \cdot \frac{\Gamma^2(k+2)}{\Gamma(2(k+2)}$

Let's see if we can simplify $\displaystyle \frac{(H_k)^2}{(k+1)^2} \cdot \frac{\Gamma^2(k+2)}{\Gamma(2(k+2)}$

$$\Gamma(k+2) = (k+1)!$$
$$\Gamma(2(k+2)) = (2k + 1)!$$

$$= \frac{(H_k)^2}{(k+1)^2} \cdot \frac{(k+1)!}{((2k+1)!}$$

Finally, $\displaystyle = (4)\cdot \sum_{k=1}^{\infty} \sum_{k=1}^{\infty}\frac{(H_k)^2}{(k+1)^2} \cdot \frac{(k+1)!}{(2k+1)!} $

I have no clue how to evaluate this sum...

 

[alyafey22]

mmm... Why can't I write the integral like that? I just factored out a $(-1)$ Is that not correct?

Ok , you should revise it because you made a small mistake.

$$= \int_{0}^{1} -2(\zeta(2) - 1) \,dx + \zeta(2)$$

There shouldn't be an integral there because you already integrated, right ?

$$= (4)\cdot\int_{0}^{1} \left( \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1}\right) \cdot \left( \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

$$= (4)\cdot \left( \sum_{k=1}^{\infty} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{(H_k)(1-x)^{k+1}}{k+1} \cdot \frac{(H_k)(x^{k+1})}{{k+1}} \right) \,dx$$

Are you sure you can do this ? I mean

$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

Let us try

$$\sum_{n \geq 0}r^n = \frac{1}{1-r} $$

Is it the case that ?

$$\sum_{n \geq 0}r^n \cdot \sum_{n \geq 0}r^n = \sum_{n \geq 0} \sum_{n \geq 0}r^{2n}$$

 

[Amad27]

Ok , you should revise it because you made a small mistake.
There shouldn't be an integral there because you already integrated, right ?
Are you sure you can do this ? I mean

$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

Let us try

$$\sum_{n \geq 0}r^n = \frac{1}{1-r} $$

Is it the case that ?

$$\sum_{n \geq 0}r^n \cdot \sum_{n \geq 0}r^n = \sum_{n \geq 0} \sum_{n \geq 0}r^{2n}$$

Question #1:

So, the answer I had for $\log(x)\log(1-x)$ is incorrect? ??

Question #2:
What are the requirements for:
$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

??

Also, mmmm. Can you give me a hint? I have the parts.

I had $\log^(1-x)$ and $\log^2(x)$ written as separate sums, but I can't take the sum because the harmonic number is there.

$\displaystyle \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $

$\displaystyle (-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1} = \log^2(x) $

Lets try the sum of $\log^2(1-x)$ first.

$\displaystyle \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$
$$\sum_{k=1}^{\infty} x^k = \frac{x}{1-x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = \int \frac{x}{1-x} \,dx$$
$$(-2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{x^{k+1}}{k+1} = (-2H_k)\cdot\int \frac{x}{1-x} \, dx = \log^2(1-x) = (2H_k)\cdot(x + \log(1-x))$$

$$\therefore \log^2(1-x) = (2H_k)\cdot(x + \log(1-x))$$

Now, let's work on $\log^2(x)$

$\displaystyle \implies \log^2(x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)((1-x)^{k+1})}{{k+1}} $

$$= \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{1} = \frac{1}{x}$$
$$= \sum_{k=1}^{\infty} \frac{(1-x)^{k}}{1} = \frac{(1-x)}{x}$$
$$= \sum_{k=1}^{\infty} (-)\cdot \frac{(1-x)^{k+1}}{k+1} = \int \frac{(1-x)}{x}$$
By part-integration, $\displaystyle (-)\cdot \sum{k=1}^{\infty} \frac{(1-x)^{k+1}}{k+1} = \log(x) - x$

Recall, $\log^(x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)((1-x)^{k+1})}{{k+1}} $

So we must multiply the previous sum by $(2H_k)$ to get:

$$\log^2(x) = (-2H_k)\cdot \sum{k=1}^{\infty} \frac{(1-x)^{k+1}}{k+1} = 2H_k(\log(x) - x)$$

Finally,

$$\log^2(x)\log^2(1-x) = \left(2H_k(\log(x) - x)\right)\cdot\left((2H_k)\cdot(x + \log(1-x))\right) = (2H_k)\left(\log(x) + \log(1-x)\right)$$

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx = (2H_k) \int_{0}^{1} \log(x) + \log(1-x) \,dx$$

Considering that $\displaystyle \int \log(u) \,du = u(\log(u) - 1)$

$$= (2H_k)\cdot(-2) = -4(H_k)$$

I don't get it. I got

$$\int_{0}^{1} \log^2(x)\log^2(1-x) \,dx = (2H_k) \int_{0}^{1} \log(x) + \log(1-x) \,dx = -4(H_k)$$

But there is no way to compute $H_k$. What can be done?

 

[alyafey22]

Question #1:

So, the answer I had for $\log(x)\log(1-x)$ is incorrect? ??

It is correct but you should have not written it like this

$$\int^1_0(-2)(\zeta(2)-1)\,dx $$

You already integrated , right ? and you got the value to be $
\zeta(2)-1$ . Why integrating again ?

Question #2:
What are the requirements for:
$$\sum a_k \cdot \sum b_k = \sum \sum a_k b_k $$

??

I must say , I don't know.

Also, mmmm. Can you give me a hint? I have the parts.

$$(-2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{x^{k+1}}{k+1} = (-2H_k)\cdot\int \frac{x}{1-x} \, dx = \log^2(1-x) = (2H_k)\cdot(x + \log(1-x))$$

No , you can't multiply both sides by $H_k$ because the value for $k$ depends on the summation , it is like you are doing that

$$\int x \,dx = \frac{x^2}{2}+c$$

Then

$$\int x^2 \,dx = \frac{x^3}{2}+c$$

which is not correct since the $x$ cannot be inserted inside the integration. The same for $H_k$ you cannot insert it inside the summation. Also when you muliply by $x$ the outer $x$ is different than the inner $x$.

I will give the hint once we agree on the above notes.

 

[Amad27]

It is correct but you should have not written it like this

$$\int^1_0(-2)(\zeta(2)-1)\,dx $$

You already integrated , right ? and you got the value to be $
\zeta(2)-1$ . Why integrating again ?
I must say , I don't know.
No , you can't multiply both sides by $H_k$ because the value for $k$ depends on the summation , it is like you are doing that

$$\int x \,dx = \frac{x^2}{2}+c$$

Then

$$\int x^2 \,dx = \frac{x^3}{2}+c$$

which is not correct since the $x$ cannot be inserted inside the integration. The same for $H_k$ you cannot insert it inside the summation. Also when you muliply by $x$ the outer $x$ is different than the inner $x$.

I will give the hint once we agree on the above notes.

Okay.

For the first integral of $\log(x)\log(1-x)$ I agree. I just realized it was a typo. I didn't pay enough attention, sorry, but I hope you know what I meant. Anyhow,

For the other harder integral, I understand. Because (H_k) is related to $k$ so we can't simply multiply it after.

I was just thinking that:

$H_k = \int_{0}^{1} \frac{1-x^k}{1-x} \,dx$

So what we had:

$\displaystyle \implies \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{{k+1}} $

$\displaystyle (-2)\cdot \sum_{k=1}^{\infty} \frac{(H_k)(1-x)^{k+1}}{k+1} = \log^2(x) $

First: $\log^2(1-x)$

$\displaystyle \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(\int_{0}^{1} \frac{1-x^k}{1-x} \,dx)(x^{k+1})}{{k+1}} $

$\displaystyle \log^2(1-x) = (-2) \cdot \sum_{k=1}^{\infty} \frac{(-\log(1-x)\cdot\int_{0}^{1} \frac{-x^k}{1-x} \,dx)(x^{k+1})}{{k+1}} $

But the integral in the numerator does not converge.

I must say, you have given an extremely tough challenge.

 

[alyafey22]

Okay , here is the hint

Only expand $\log^2(1-x)$ as series. Leave $\log^2(x)$ as it is.

You will see later , why!

 

[Amad27]

Okay , here is the hint

Only expand $\log^2(1-x)$ as series. Leave $\log^2(x)$ as it is.

You will see later , why!

Hello ZaidAlyafey,

Lets see what we got here:

$$\log^2(1-x) = (-2)\cdot\sum_{k=1}^{\infty}\frac{(H_k)(x)^{(k+1)}}{(k+1)}$$

$$\log^2(1-x)\log^2(x) = (2)\cdot\sum_{k=1}^{\infty} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)}$$

$$\int_{0}^{1}\log^2(1-x)\log^2(x) \,dx = (2)\cdot \int_{0}^{1} \sum_{k=1}^{\infty} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)} \,dx$$

$$\int_{0}^{1} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)} \,dx = \frac{H_k}{k+1}\cdot \int_{0}^{1} \log^2(x)\cdot (x)^{k+1} \,dx$$

The integral is the tough part.

It is very well displayed by the second partial derivative of the beta function but I yet have no clue how to compute trigammas etc.

Through integration by parts I was able to do it.

$$\int_{0}^{1} \frac{(H_k)\log^2(x)(x)^{(k+1)}}{(k+1)} \,dx = \frac{H_k}{k+1}\cdot \int_{0}^{1} \log^2(x)\cdot (x)^{k+1} \,dx = \frac{H_k}{k+1} \cdot \frac{(2)}{(k+2)^3}$$

So we have,

$$I = \sum_{k=1}^{\infty}\frac{2H_k}{(k+1)(k+2)^3}$$

What is left is, is to find this difficult sum. It is tough because of the harmonic number...

I'm not sure...

$H_k = \sum_{n=1}^{k} \frac{1}{n}$

This is quite difficult =)

 

[alyafey22]

$$I = \sum_{k=1}^{\infty}\frac{2H_k}{(k+1)(k+2)^3}$$

Very good , nice work! Now we should use partial fractions , right ?

 

[Amad27]

Very good , nice work! Now we should use partial fractions , right ?

That was one of my attempts, but it doesn't work, I'll try again here:

The harmonic number is not a constant. Partial fraction requires the numerator to be a constant! Ah - now I see why I was failing miserably.

 

[alyafey22]

And you may use the following

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

If looking for a proof .

 

[Amad27]

And you may use the following

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

If looking for a proof .

Thanks for the hint, but I don't think it will apply in this case.

The denominator is not one variables, it is a product of two binomials, which the theorem you stated does not apply for.

 

[alyafey22]

Thanks for the hint, but I don't think it will apply in this case.

The denominator is not one variables, it is a product of two binomials, which the theorem you stated does not apply for.

That's why you have to use partial fractions.

 

[Amad27]

That's why you have to use partial fractions.

We had the following,

$$\frac{2H_k}{(k+1)(k+2)^3}$$

$$\frac{A}{k+1}+\frac{B}{k+2}+\frac{C}{(k+2)^2}+\frac{D}{(k+2)^3} = \frac{2H_k}{(k+1)(k+2)^3}$$

$$(A)(k+2)^3 + (k+1)(k+2)^2(B) + (k+1)(k+2)(C) + (k+1)(D) = 2H_k$$

I was thinking $k=-1$ but I am not sure if $H_{-1}$ exists.

 

[alyafey22]

We had the following,

$$\frac{2H_k}{(k+1)(k+2)^3}$$

$$\frac{A}{k+1}+\frac{B}{k+2}+\frac{C}{(k+2)^2}+\frac{D}{(k+2)^3} = \frac{2H_k}{(k+1)(k+2)^3}$$

$$(A)(k+2)^3 + (k+1)(k+2)^2(B) + (k+1)(k+2)(C) + (k+1)(D) = 2H_k$$

I was thinking $k=-1$ but I am not sure if $H_{-1}$ exists.

Ignore $H_k$ , just find the partial fraction decomposition for

$$\frac{1}{(k+1)(k+2)^3}$$

 

[Amad27]

Ignore $H_k$ , just find the partial fraction decomposition for

$$\frac{1}{(k+1)(k+2)^3}$$

That is simply, (skipping steps):

$$\frac{1}{(k+1)(k+2)^3} = -(\frac{1}{k+2} + \frac{1}{(k+2)^2} + \frac{1}{(k+2)^3}) + \frac{1}{k+1}$$

$$\frac{2H_k}{(k+1)(k+2)^3} = \frac{-2H_k}{k+2} - \frac{2H_k}{(k+2)^2} - \frac{2H_k}{(k+2)^3} + \frac{2H_k}{k+1}$$

But the theorem you stated still won't apply as the denominator is still a binomial...

 

[alyafey22]

Hint

$$\sum_{k=1}^\infty \frac{H_k}{(k+1)^q} = \sum_{n=2}^\infty \frac{H_{n-1}}{n^q} = \sum_{n=2}^\infty \frac{H_n -\frac{1}{n}}{n^q} = \sum_{n=2}^\infty\frac{H_n}{n^{q+1}}-(\zeta(q+1)-1)$$

You can use the same concept to find

$$\sum_{k=1}^\infty \frac{H_k}{(k+2)^q}$$