Challenging Logarithmic Integral: What Methods Can Solve It?

[Amad27]

Hello,

The problem is

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

Is it even worth trying?

I haven't learned series yet, so if there are other methods, please let me know so I can start off. I did some differentiation under the integral sign, but it didnt work out well.

If anyone wants to see the approach I'll post it. Ideas?

Thanks!

 

[Euge]

Hello,

The problem is

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

Is it even worth trying?

I haven't learned series yet, so if there are other methods, please let me know so I can start off. I did some differentiation under the integral sign, but it didnt work out well.

If anyone wants to see the approach I'll post it. Ideas?

Thanks!

Hi Olok,

I say this respectfully, but if you haven't learned series yet, you're not likely going to understand this problem even by method of differentiation under the integral sign. Regardless, here's a possible way to approach the problem. Consider

$$ F(a,b,c) \stackrel{\text{def}}= \int_0^1 (1 - x)^a (1 + x)^b (1 + 2x)^c\, dx.$$

Using the differentiation rule $\frac{\partial}{\partial u} w^u = w^u \ln u$, we find that

$$ \int_0^1 \frac{\ln(1 - x) \ln(1 + x) \ln(1 + 2x)}{1 + 2x}\, dx = \lim_{{{a \to 0 \atop b \to 0}\atop c \to -1}} \frac{\partial^3 F}{\partial a\, \partial b\, \partial c}(a,b,c).$$

You will need to express $F(a,b,c)$ in terms of some special functions before computing third derivatives.

 

[Amad27]

Hi Olok,

I say this respectfully, but if you haven't learned series yet, you're not likely going to understand this problem even by method of differentiation under the integral sign. Regardless, here's a possible way to approach the problem. Consider

$$ F(a,b,c) \stackrel{\text{def}}= \int_0^1 (1 - x)^a (1 + x)^b (1 + 2x)^c\, dx.$$

Using the differentiation rule $\frac{\partial}{\partial u} w^u = w^u \ln u$, we find that

$$ \int_0^1 \frac{\ln(1 - x) \ln(1 + x) \ln(1 + 2x)}{1 + 2x}\, dx = \lim_{{{a \to 0 \atop b \to 0}\atop c \to -1}} \frac{\partial^3 F}{\partial a\, \partial b\, \partial c}(a,b,c).$$

You will need to express $F(a,b,c)$ in terms of some special functions before computing third derivatives.

Hello @Euge,

Unfortunately. I am still in high school, with less time to explore anything on my own. But I am very curious how you can use series for evaluating integrals.

Let's take a simple example.

$\int_{0}^{1} e^{2x} \,dx $

Can you quickly show me how you can use series to evaluate that specific integral? Just quickly, so I can get a feel for it =)

Thanks!

 

[Euge]

Hello @Euge,

Unfortunately. I am still in high school, with less time to explore anything on my own. But I am very curious how you can use series for evaluating integrals.

Let's take a simple example.

$\int_{0}^{1} e^{2x} \,dx $

Can you quickly show me how you can use series to evaluate that specific integral? Just quickly, so I can get a feel for it =)

Thanks!

Ok, if it helps!

Since $$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$ for all real $x$, we have

$$\int_0^1 e^{2x}\, dx = \int_0^1 \sum_{n = 0}^\infty \frac{(2x)^n}{n!} = \sum_{n = 0}^\infty \frac{2^n}{n!} \int_0^1 x^n\, dx = \sum_{n = 0}^\infty \frac{2^n}{n!} \cdot \frac{1}{n+1}$$

$$ = \sum_{n = 0}^\infty \frac{2^n}{(n+1)!} = \frac{1}{2} \sum_{n = 0}^\infty \frac{2^{n+1}}{(n+1)!} = \frac{1}{2} \sum_{n = 1}^\infty \frac{2^n}{n!} = \frac{1}{2} (e^2 - 1).$$

The justification for interchanging the series and integral is that power series may be integrated term-by-term within its interval of convergence (in this example, $$e^{2x}$$ has an infinite radius of convergence).

 

[Amad27]

Ok, if it helps!

Since $$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$ for all real $x$, we have

$$\int_0^1 e^{2x}\, dx = \int_0^1 \sum_{n = 0}^\infty \frac{(2x)^n}{n!} = \sum_{n = 0}^\infty \frac{2^n}{n!} \int_0^1 x^n\, dx = \sum_{n = 0}^\infty \frac{2^n}{n!} \cdot \frac{1}{n+1}$$

$$ = \sum_{n = 0}^\infty \frac{2^n}{(n+1)!} = \frac{1}{2} \sum_{n = 0}^\infty \frac{2^{n+1}}{(n+1)!} = \frac{1}{2} \sum_{n = 1}^\infty \frac{2^n}{n!} = \frac{1}{2} (e^2 - 1).$$

The justification for interchanging the series and integral is that power series may be integrated term-by-term within its interval of convergence (in this example, $$e^{2x}$$ has an infinite radius of convergence).

Hello,

I might be a bit of a "non-knower" in this part but how is

$\sum_{0}^{\infty} \frac{2^n}{n!}$

This part a constant? How is this a constant that you can just pull out? In a simple question,

How is that expression a constant?

Thanks!

 

[Euge]

The expression doesn't depend on $x$. If that's not satisfying enough, plug in $x = 2$ in the formula

$$ e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}.$$

Doing so, you find that the value of the series in question is $e^2$, a constant.

 

[Amad27]

The expression doesn't depend on $x$. If that's not satisfying enough, plug in $x = 2$ in the formula

$$ e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}.$$

Doing so, you find that the value of the series in question is $e^2$, a constant.

Oh, so I have an idea.

A constant is one you cannot manipulate.

$x$ you can manipulate, it can change. Do you consider that a constant (the sum) because it is a sum??

That a sum is definite?

Thanks!

 

[Euge]

By definition a constant does not change value.

 

[Amad27]

By definition a constant does not change value.

And the sum was a constant, because it was definite?

 

[mathbalarka]

I am not sure what you mean by "definite". A constant is usually defined as a complex number. In this case, the sum

$$\sum_{n = 1}^\infty \frac{2^n}{n!}$$

converges (can you prove it). Thus, it is an element of $\Bbb C$.

 

[Amad27]

I am not sure what you mean by "definite". A constant is usually defined as a complex number. In this case, the sum

$$\sum_{n = 1}^\infty \frac{2^n}{n!}$$

converges (can you prove it). Thus, it is an element of $\Bbb C$.

I meant something similar.

@Euge pulled it out as considering it a constant.

I reasoned that,

$\sum_{n = 1}^\infty \frac{2^n}{n!}$ is a constant, because,

It is "definite." Meaning that that sum is one single number. As you said, it converges, so it is a single number.

 

[Euge]

And the sum was a constant, because it was definite?

The series converges by the ratio test. Indeed, since

$$\dfrac{\left|\frac{2^{n+1}}{(n+1)!}\right|}{\left|\frac{2^n}{n!}\right|} = \frac{2}{n+1} \to 0$$ as $$n \to \infty$$,

the ratio test ensures that the series $$\sum_{n = 0}^\infty \frac{2^n}{n!}$$ converges.

 

[Amad27]

The series converges by the ratio test. Indeed, since

$$\dfrac{\left|\frac{2^{n+1}}{(n+1)!}\right|}{\left|\frac{2^n}{n!}\right|} = \frac{2}{n+1} \to 0$$ as $$n \to \infty$$,

the ratio test ensures that the series $$\sum_{n = 0}^\infty \frac{2^n}{n!}$$ converges.

Nice, so you have proved it converges.

Does that prove it is a constant? Yes, right?

I'm sounding a bit obnoxious because I haven't had a formal introduction to sequences and series yet -_- =D

 

[mathbalarka]

Whether it is a constant or not is irrelevant of pulling it out from the integral. Consider

$$F(t) = \int_a^b t \cdot \sqrt{x} \, \mathrm{d}x$$

$F(t)$ is a function of $t$, thus $t$ most definitely isn't a constant. But as $t$ is independent of the argument $x$ of the integral, you can pull it out from the integral :

$$\int_a^b t \cdot \sqrt{x} \, \mathrm{d}x = t \int_a^b \sqrt{x} \, \mathrm{d}x$$

This is just another form of the distributive property of addition and multiplication : $$\begin{align}\sum_{n = a}^b f(t) g(n) &= f(t)g(a) + f(t)g(a+1) + \cdots + f(t)g(b) \\ &= f(t) \left [ g(a) + g(a+1) + \cdots + g(b) \right ] \\ &= f(t) \sum_{n = a}^b g(n)\end{align}$$

 

[Amad27]

Whether it is a constant or not is irrelevant of pulling it out from the integral. Consider

$$F(t) = \int_a^b t \cdot \sqrt{x} \, \mathrm{d}x$$

$F(t)$ is a function of $t$, thus $t$ most definitely isn't a constant. But as $t$ is independent of the argument $x$ of the integral, you can pull it out from the integral :

$$\int_a^b t \cdot \sqrt{x} \, \mathrm{d}x = t \int_a^b \sqrt{x} \, \mathrm{d}x$$

This is just another form of the distributive property of addition and multiplication : $$\begin{align}\sum_{n = a}^b f(t) g(n) &= f(t)g(a) + f(t)g(a+1) + \cdots + f(t)g(b) \\ &= f(t) \left [ g(a) + g(a+1) + \cdots + g(b) \right ] \\ &= f(t) \sum_{n = a}^b g(n)\end{align}$$

Wow, thanks @mathbalarka, that was an amazing explanation, I'd like to elaborate on that.

$\begin{align}\sum_{n = a}^b f(t) g(n) &= f(t)g(a) + f(t)g(a+1) + \cdots + f(t)g(b) \\ &= f(t) \left [ g(a) + g(a+1) + \cdots + g(b) \right ] \\ &= f(t) \sum_{n = a}^b g(n)\end{align}$

The "star" of my elaboration is, (I tweaked it for simplicity)

$f(t) \cdot \sum_{k = 1}^{n} g(x)$

$\lim_{{\left\lVert{\Delta}\right\rVert}\to{0}} f(t) \cdot \sum_{k = 1}^{n} g(x)$

Let there be $n$ subintervals (unequal length).

$f(t) \cdot \int_{a}^{b} g(x) \,dx$

 

[Amad27]

Whether it is a constant or not is irrelevant of pulling it out from the integral. Consider

$$F(t) = \int_a^b t \cdot \sqrt{x} \, \mathrm{d}x$$

$F(t)$ is a function of $t$, thus $t$ most definitely isn't a constant. But as $t$ is independent of the argument $x$ of the integral, you can pull it out from the integral :

$$\int_a^b t \cdot \sqrt{x} \, \mathrm{d}x = t \int_a^b \sqrt{x} \, \mathrm{d}x$$

This is just another form of the distributive property of addition and multiplication : $$\begin{align}\sum_{n = a}^b f(t) g(n) &= f(t)g(a) + f(t)g(a+1) + \cdots + f(t)g(b) \\ &= f(t) \left [ g(a) + g(a+1) + \cdots + g(b) \right ] \\ &= f(t) \sum_{n = a}^b g(n)\end{align}$$

Hello,

If you notice my latest comment, my summation nor the integral came out properly.

The summation has the index in the wrong place. How can I format the summation and the integral properly?

Thanks

 

[mathbalarka]

Indeed, the correct Riemann sum should have been

$$\sum_{a \leq i < b} f(t)g(x'_i) \Delta x_i$$

For $\Delta x_i = x_{i+1} - x_i$ and $x'_i \in [x_{i+1}, x_i]$. Hence note that

$$\begin{align}\sum_{a \leq i < b} f(t) g(x'_i) \Delta x_i &= f(t) g(x'_a) \Delta x_a + f(t) g(x'_{a+1}) \Delta x_{a+1} + \cdots + f(t) g(x'_{b-1}) \Delta x_{b-1} \\ & = f(t)[g(x'_a)\Delta x_a + f(x'_{a+1})\Delta x_{a+1} + \cdots + f(x'_{b-1})\Delta x_{b-1}] \\& = f(t)\sum_{a \leq i < b} g(x'_i) \Delta x_i \end{align}$$

Thus let $\Delta x_i \to 0$ to get

$$\int_a^b f(t)g(x) dx = f(t) \int_a^b g(x) dx$$

 

[Amad27]

Indeed, the correct Riemann sum should have been

$$\sum_{a \leq i < b} f(t)g(x'_i) \Delta x_i$$

For $\Delta x_i = x_{i+1} - x_i$ and $x'_i \in [x_{i+1}, x_i]$. Hence note that

$$\begin{align}\sum_{a \leq i < b} f(t) g(x'_i) \Delta x_i &= f(t) g(x'_a) \Delta x_a + f(t) g(x'_{a+1}) \Delta x_{a+1} + \cdots + f(t) g(x'_{b-1}) \Delta x_{b-1} \\ & = f(t)[g(x'_a)\Delta x_a + f(x'_{a+1})\Delta x_{a+1} + \cdots + f(x'_{b-1})\Delta x_{b-1}] \\& = f(t)\sum_{a \leq i < b} g(x'_i) \Delta x_i \end{align}$$

Thus let $\Delta x_i \to 0$ to get

$$\int_a^b f(t)g(x) dx = f(t) \int_a^b g(x) dx$$

A question and a comment.

comment first

I never seen that form of a Riemann-sum, I think what you have written might be related to measure theory. Interesting though. Question second

So when the part is unrelated it can be pulled out of the integral?

$\int_{a}^{b}f(x) t \,dx$

Implicitly, we define a variable $x$, and $t$

What is $t$?

Is $t$ a variable? Or
Is $t$ a constant?

What is it?

Thanks!

 

[mathbalarka]

I never seen that form of a Riemann-sum, I think what you have written might be related to measure theory. Interesting though.

I think you are having terminological issues. The standard form of Riemann sum for $F$ defined on $[a, b]$ is

$$\sum_{a \leq n < b} F(x'_i) \Delta x_i$$

And I have used this for $F(x) = f(t)g(x)$, $t$ some variable independent of $x$.

What is t? Is t a variable? Or Is t a constant?

$t$ is either a constant or a variable independent of $x$, i.e., it's a constant "relative" to $x$.

 

[Amad27]

I think you are having terminological issues. The standard form of Riemann sum for $F$ defined on $[a, b]$ is

$$\sum_{a \leq n < b} F(x'_i) \Delta x_i$$

And I have used this for $F(x) = f(t)g(x)$, $t$ some variable independent of $x$.
$t$ is either a constant or a variable independent of $x$, i.e., it's a constant "relative" to $x$.

Hello @mathbalarka.

When you say

$x'_i$

I have never learned what this $'$ means in this. I've seen that apostrophe several times, but I am not sure what the definition of that is.

second part,

Is there a theorem or definition, which perhaps states if

$t$ is either a constant or variable unrelated; independent to $x$ then it can be pulled out of an integral expression?

Thanks!

 

[mathbalarka]

Hello @mathbalarka.

When you say

$x'_i$

I have never learned what this $'$ means in this. I've seen that apostrophe several times, but I am not sure what the definition of that is.

$x'_i$ is a point inside $[x_i, x_{i+1}]$, I think I have mentioned it before.

Is there a theorem or definition, which perhaps states if

$t$ is either a constant or variable unrelated; independent to $x$ then it can be pulled out of an integral expression?

I thought I had proved it above? :

$$\sum_{a \leq i < b} f(t)g(x'_i) \Delta x_i$$

For $\Delta x_i = x_{i+1} - x_i$ and $x'_i \in [x_{i+1}, x_i]$. Hence note that

$$\begin{align}\sum_{a \leq i < b} f(t) g(x'_i) \Delta x_i &= f(t) g(x'_a) \Delta x_a + f(t) g(x'_{a+1}) \Delta x_{a+1} + \cdots + f(t) g(x'_{b-1}) \Delta x_{b-1} \\ & = f(t)[g(x'_a)\Delta x_a + f(x'_{a+1})\Delta x_{a+1} + \cdots + f(x'_{b-1})\Delta x_{b-1}] \\& = f(t)\sum_{a \leq i < b} g(x'_i) \Delta x_i \end{align}$$

Thus let $\Delta x_i \to 0$ to get

$$\int_a^b f(t)g(x) dx = f(t) \int_a^b g(x) dx$$

 

[Amad27]

$x'_i$ is a point inside $[x_i, x_{i+1}]$, I think I have mentioned it before.
I thought I had proved it above? :

It seems quite absurd, but I suppose that is how it is.

Pulling out a variable like $t$ is something new.

I suppose it it something new I've learned today!

Thanks =)