MHB Integration using Beta and Gamma Functions

[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

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

 

[Amad27]

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

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

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

 

[alyafey22]

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

I cannot follow , could you rewrite that!

 

[Amad27]

I cannot follow , could you rewrite that!

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}$

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

 

[alyafey22]

It follows that:

$H_k = H_{k-1} + \frac{1}{k}$

From the original hint:

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

$H_{k-1} = H_k + \frac{1}{k}$

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

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

Let n = k-1

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

Is this the derivation? and then we split up the numerator to get the other formula.

And I suppose we use the same method from $k+2$

Yes , but remember that in our problem $k$ starts from 1 not 2 , so you have to take care of that.

By the way, you sent me the URL:
proof .

It didnt show up. I'll try to come up with my own proof (so don't show a link yet!)

Thanks =)

Ok , good luck with a nice proof.

 

[Amad27]

Yes , but remember that in our problem $k$ starts from 1 not 2 , so you have to take care of that.

Ok , good luck with a nice proof.

FIRST QUESTION:
Where do you find such formulas in the first place? I mean, I tried searching online but couldn't find traces of that formula, so how did you discover it and where?

NEXT:

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

We know: $H_k = H_{k-1} + \frac{1}{k}$

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

Let $n = k + 1$, the starting point is then: $k = 0$

$$\sum_{k=0}^\infty \frac{H_{k} + \frac{1}{(k+1)}}{(k+1)^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z)$$

$$\sum_{k=0}^\infty \frac{H_{k} + \frac{1}{(k+1)}}{(k+1)^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z)$$

$$\sum_{k=0}^\infty \frac{H_{k}}{(k+1)^q} + \frac{1}{(k+1)^{q+1}}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z)$$

$$\therefore \sum_{k=1}^\infty \frac{H_{k}}{(k+1)^q} = \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{z=1}^{q-2}\zeta(z+1)\zeta(q-z) -\sum_{k=0}^\infty \frac{1}{(k+1)^{q+1}}$$

Let $q = 1$

$$= (3/2)\zeta(2) - (1/2)(\sum_{z=1}^{-1} \zeta(z+1)\zeta(0)) - \sum_{k=0}^{\infty}\frac{1}{(k+1)^{q+1}}$$

There is an issue here:

$$(\sum_{z=1}^{-1} \zeta(z+1)\zeta(0))$$ does not exist.

 

[alyafey22]

The following sum diverges

$$\sum_{k\geq 1}\frac{H_k}{k}$$

Actually the Euler number $\gamma$ can be defined as

$$\gamma = \lim_{k\to \infty}H_k-\log(k)$$

So Harmonic number differ by a constant term as they approach infinity.

Hence we can write

$$\sum_{k\geq 1}\frac{H_k}{k} \sim \sum_{k\geq 1}\frac{\log(k)}{k} $$

The right integral diverges by the integral test.

You don't have to understand all of that but you have to know that diverges.

So the equation

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

only converges for $q \geq 2$ and by definition for $q=2$ we have

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(2-k)$$

The sum is by definition equal to 0 so

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)$$

For references you can search for Euler sums , you will find lots of papers.

 

[Amad27]

The following sum diverges

$$\sum_{k\geq 1}\frac{H_k}{k}$$

Actually the Euler number $\gamma$ can be defined as

$$\gamma = \lim_{k\to \infty}H_k-\log(k)$$

So Harmonic number differ by a constant term as they approach infinity.

Hence we can write

$$\sum_{k\geq 1}\frac{H_k}{k} \sim \sum_{k\geq 1}\frac{\log(k)}{k} $$

The right integral diverges by the integral test.

You don't have to understand all of that but you have to know that diverges.

So the equation

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

only converges for $q \geq 2$ and by definition for $q=2$ we have

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(2-k)$$

The sum is by definition equal to 0 so

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)$$

For references you can search for Euler sums , you will find lots of papers.

But the issue is we don't have an $n^2$ term when evaluating the $k+1$ sum. Please let me know if there is a work around for this.
Meanwhilw, I just had an idea.

We can consider the integral definition of the Harmonix number.

$$H_k = \int_{0}^{1} (1-x)^{k-1} dx$$

And convert the whole sum.

I used integral on the original geometric series formula and I got an intense formula involving several poly logarithms. Perhaps this is a way? Will you be interested on seeing that attempt?

 

[alyafey22]

But the issue is we don't have an $n^2$ term when evaluating the $k+1$ sum. Please let me know if there is a work around for this.

I don't understand what you mean .

Meanwhilw, I just had an idea.

We can consider the integral definition of the Harmonix number.

$$H_k = \int_{0}^{1} (1-x)^{k-1} dx$$

And convert the whole sum.

I used integral on the original geometric series formula and I got an intense formula involving several poly logarithms. Perhaps this is a way? Will you be interested on seeing that attempt?

Of course , let me see if you got any other attempts. It might be better than my approach , why not ?

 

[Amad27]

I don't understand what you mean .
Of course , let me see if you got any other attempts. It might be better than my approach , why not ?

So we consider just

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

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

Now it is about to get extremely complicated. Principally, we know

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

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

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

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

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

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

I need help here:

I can't integrate because I checked on W|A and there is a dilogarithmic term involved.

How do you express polylogs as indefnite integrals? So that there is no upper or lower bound?

From what I know:

$$Li_2(x) = (-)\cdot\int \frac{\log(x)}{(1-x)}$$
$$Li_2(1-x) = (-)\cdot \int \frac{\log(1-x)}{x}$$

Perhaps we could use integration by parts to get this factor in there?

There is always a question I've had:

How can you define a function $f(a)$ such that $a$ is actually inside the integral? So:

$$f(a) = \int ax^2 dx$$

Usually, textbooks define integral functions as the variable being the upper limit of a definite integral.

So what property/rule of integral, or definition states you can have a function of a variable inside the integral itself? Thanks!

 

[alyafey22]

$$H_k = \int_{0}^{1} (1-x)^{k-1} dx$$

Sorry , I missed the definition ... Why is that true ?

 

[Amad27]

Sorry , I missed the definition ... Why is that true ?

Oh oh...

That's why it wasn't working.

I used the wrong definition. It is

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

The $(1-x)$ isn't in brackets...

Please give me some time to think over this. Meanwhile... can you answer the other question I asked?

Question:
How can you define a function $f(a)$ such that $a$ is actually inside the integral? So:

$$f(a) = \int ax^2 dx$$

Usually, textbooks define integral functions as the variable being the upper limit of a definite integral.

So what property/rule of integral, or definition states you can have a function of a variable inside the integral itself? Thanks!

 

[alyafey22]

Question:
How can you define a function $f(a)$ such that $a$ is actually inside the integral? So:

$$f(a) = \int ax^2 dx$$

Usually, textbooks define integral functions as the variable being the upper limit of a definite integral.

So what property/rule of integral, or definition states you can have a function of a variable inside the integral itself? Thanks!

You can do the following

$$f(a) = \int^{t_1}_{t_0} g(t,a)\,dt$$

where $t_0,t_1$ are constants.

Indeed , it is just a symbol to define the integral. For example we have

$$\Gamma(a) = \int^\infty_0 t^{a-1} \, e^{-t} \, dt$$

where in this case $f(a) = \Gamma(a) $ and $g(t,a) = t^{a-1} \, e^{-t} $.

Since $a$ is independent of $t$ we can define something like this

$$g(a) = a \int^1_0 f(t) \, dt $$

I don't recall a name for this property but is mainly used to differentiate w.r.t to the variable inserted.

 

[Amad27]

You can do the following

$$f(a) = \int^{t_1}_{t_0} g(t,a)\,dt$$

where $t_0,t_1$ are constants.

Indeed , it is just a symbol to define the integral. For example we have

$$\Gamma(a) = \int^\infty_0 t^{a-1} \, e^{-t} \, dt$$

where in this case $f(a) = \Gamma(a) $ and $g(t,a) = t^{a-1} \, e^{-t} $.

Since $a$ is independent of $t$ we can define something like this

$$g(a) = a \int^1_0 f(t) \, dt $$

I don't recall a name for this property but is mainly used to differentiate w.r.t to the variable inserted.

I suppose it is a property of integral functions. Anywho.

I worked on the problem, and found out how to do it... until a point.

We understand

$$I = \int_{0}^{1} \left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} + \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^2}\right) \, dx$$

Assuming both of these sums are convergent. I used the formula,

$$\sum_{k=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$ to get a formula for a general $x^{k+2}$.

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_3(x) - Li_2(x) - \frac{x^2}{8} + 3x - x\log(1-x) + \log(1-x)$$

If you want to see the workings, I can try to scan it, because it took a few pages... (I used W|A for some integrals, instead of doing integration by parts).

Letting $x=1$, you get

$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} = -Li_3(1) - Li_2(1) - \frac{1}{8} + 3 - \log(0) + \log(0)$$

Here is the crises, $\log(0)$ is undefined, is it okay to subtract this: $\text{undefined} - \text{undefined}$ ?? Let me know of this property/theorem.

$$= -\zeta(3) - \zeta(2) + \frac{23}{8}$$

I am working on the other part, but was the logarithmic subtraction correct?

Thanks

 

[alyafey22]

$$I = \int_{0}^{1} \left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} + \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^2}\right) \, dx$$

I don't seem to understand how you got that!

You used

$$H_k = \int^1_0 \frac{1-x^k}{1-x}\,dx $$

right ?

 

[Amad27]

I don't seem to understand how you got that!

You used

$$H_k = \int^1_0 \frac{1-x^k}{1-x}\,dx $$

right ?

FIRST A QUESTION:
I asked this before, but can we do $\log(0) - \log(0) = 0$? Otherwise this idea doesn't work.

Now, the workings:

I'll try to do the best I can.

$$H_k = \int^1_0 \frac{1-x^k}{1-x}\,dx $$
We had

$$(2)\cdot\int_{0}^{1} \frac{1}{1-x}\left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} - \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^3}\right) \,dx$$

Through Fubini's theorem, you interchange the sum and integral, then $1/(1-x)$ is just a constant for the sum, so you can "extract," that out. Then split the sum, ASSUMING both are convergent. Since I am too lazy to do a test. Actually I don't know how to test a series for convergence anyway...

Then the sum of

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -[Li_3(x) + Li_2(x)] - x^2/8 + 3x - x\log(1-x) + \log(1-x)$$

Letting $x=1$, the sum is: $S_1 = -\zeta(3) - \zeta(2) + 23/8$

Dividing the sum of $x^{k+2}$ by $x^2$ we get.

$$\sum_{k=1}^{\infty} \frac{x^{k}}{(k+1)(k+2)^3} = -[Li_3(x) + Li_2(x)]/x^2 - [x^2/8 + 3x - x\log(1-x) + \log(1-x)]/x^2$$

We had

$$(2)\cdot\int_{0}^{1} \frac{1}{1-x}\left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} - \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^3}\right) \,dx$$

$$= (2)\cdot\int_{0}^{1} \frac{1}{1-x}\left(-\zeta(3) - \zeta(2) + 23/8 - \left(-[Li_3(x) + Li_2(x)]/x^2 - [x^2/8 + 3x - x\log(1-x) + \log(1-x)]/x^2\right)\right) \,dx$$

$$= (2)\cdot\int_{0}^{1} \frac{-\zeta(3) - \zeta(2) + 23/8}{1-x}\left( - \left(\frac{-Li_3(x) - Li_2(x) - \frac{x^2}{8} - 3x + x\log(1-x) - \log(1-x)}{(1-x)x^2}\right)\right) \,dx$$

$$ = (2)\cdot\int_{0}^{1} \frac{-x^2\zeta(3) - x^2\zeta(2) + \frac{23x^2}{8} + Li_3(x) + Li_2(x) + \frac{x^2}{8} + 3x - x\log(1-x) + \log(1-x)}{(1-x)x^2} \,dx$$

I think we should split this, but this integral product should give the final answer.

But how come $\displaystyle \int_{0}^{1} \frac{Li_3(x) + Li_2(x)}{(1-x)x^2} \,dx$ does NOT converge?

 

[alyafey22]

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -[Li_3(x) + Li_2(x)] - x^2/8 + 3x - x\log(1-x) + \log(1-x)$$

Let us go step by step first. How did you prove the above formula ?

 

[Amad27]

Let us go step by step first. How did you prove the above formula ?

I will post it the second I get home. But for now,

How can I check if an integral converges or not?

I tried the integral

$$\int_{0}^{1} \frac{-Li_3(x) - Li_2(x)}{(1-x)x^2} \,dx$$

But it does not converge. How can I know if the integral, will converge or not BEFOREhand.QUESTION 2: Which types of integrals involving Polylogs converge?

Thanks =)

 

[alyafey22]

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

Does not converge. This can be see using

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

Then

$$\mathrm{Li}_2(1-x) = \zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx = \int^1_0 \frac{\zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)}{x}\,dx $$

First integral

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

The following integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx$$

converges since , function to integrate is continuous so the integral converges.

It remains

$$\int^1_0 \frac{\zeta(2)}{x}\,dx$$

is clearly divergent.

There are many ways to apply , but generally you have to do some change of variable or integration by part to see why the integral diverges. Sometimes , you have to do some computations.

In other cases , you can bound the integral. For example

$$\int^\infty_1 \frac{e^{-x}}{x^2}\,dx$$

How to prove that the improper integral converges ?

$$|\int^\infty_1 \frac{e^{-x}}{x^2}\,dx| \leq \int^\infty_1 | \frac{e^{-x}}{x^2}| \,dx \leq \int^\infty_1 \frac{1}{x^2} < \infty$$

Since the area of the integral is less than a finite area , the integral converges.

 

[Amad27]

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

Does not converge. This can be see using

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

Then

$$\mathrm{Li}_2(1-x) = \zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx = \int^1_0 \frac{\zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)}{x}\,dx $$

First integral

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

The following integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx$$

converges since , function to integrate is continuous so the integral converges.

It remains

$$\int^1_0 \frac{\zeta(2)}{x}\,dx$$

is clearly divergent.

There are many ways to apply , but generally you have to do some change of variable or integration by part to see why the integral diverges. Sometimes , you have to do some computations.

In other cases , you can bound the integral. For example

$$\int^\infty_1 \frac{e^{-x}}{x^2}\,dx$$

How to prove that the improper integral converges ?

$$|\int^\infty_1 \frac{e^{-x}}{x^2}\,dx| \leq \int^\infty_1 | \frac{e^{-x}}{x^2}| \,dx \leq \int^\infty_1 \frac{1}{x^2} < \infty$$

Since the area of the integral is less than a finite area , the integral converges.

Here's the proof

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$
$$\sum_{k=1}^{\infty} x^{k} = \frac{x}{1-x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = -x - \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)} = \frac{-x^2}{2} + x - x\log(1-x) + \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)} = \frac{-x}{2} + 1 - \log(1-x) + \frac{\log(1-x)}{x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^2} = \log(1-x) - x\log(1-x) + 2x - \frac{x^2}{4} - Li_2(x) $$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)^2} = \frac{\log(1-x)}{x} - \log(1-x) + 2 - \frac{x}{4} - \frac{Li_2(x)}{x} $$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x) \blacksquare$$

 

[alyafey22]

Here's the proof

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$
$$\sum_{k=1}^{\infty} x^{k} = \frac{x}{1-x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = -x - \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)} = \frac{-x^2}{2} + x - x\log(1-x) + \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)} = \frac{-x}{2} + 1 - \log(1-x) + \frac{\log(1-x)}{x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^2} = \log(1-x) - x\log(1-x) + 2x - \frac{x^2}{4} - Li_2(x) $$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)^2} = \frac{\log(1-x)}{x} - \log(1-x) + 2 - \frac{x}{4} - \frac{Li_2(x)}{x} $$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x) \blacksquare$$

Seems correct. Let us verify it by taking the limit as $x \to 1$.

What do you get ?

Does it agree with the value returned by wolfram.