# Challenge 20: Pranav-Arora's Integral

1. Jul 22, 2014

### micromass

Pranav-Arora has sent me an excellent math challenge for this week. The problem statement is easy:

Please list any sources that you have used to solve this question. Using google or other search engines is forbidden. Wikipedia is allowed but not its search engine. Wolfram alpha and other mathematical software are not allowed.

2. Jul 22, 2014

### pasmith

The obvious substitution is $x = e^u$, yielding $$I = \int_0^\infty \frac{x^3 (\ln x)^2}{1 + x^6}\,dx = \int_{-\infty}^\infty \frac{u^2 e^u}{e^{3u} + e^{-3u}}\,du \\ = \int_0^\infty \frac{u^2 e^u}{e^{3u} + e^{-3u}}\,du + \int_0^\infty \frac{u^2 e^{-u}}{e^{3u} + e^{-3u}}\,du$$ by evenness of $u^2/ (e^{3u} + e^{-3u})$. Setting
$$F(k) = \int_0^\infty \frac{u^2 e^{ku}}{e^{3u} + e^{-3u}}\,du$$ we have that $I = F(1) + F(-1)$.

Writing $$(e^{3u} + e^{-3u})^{-1} = e^{-3u}(1 + e^{-6u})^{-1}$$ we can expand in binomial series to obtain $$F(k) = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^2 e^{ku} e^{-3u} e^{-6nu}\,du = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^2 e^{-u(6n + 3 - k)}\,du$$ and substituting $v = u(6n + 3 - k)$ yields $$\int_0^\infty u^2 e^{-u(6n + 3 - k)}\,du = \frac{1}{(6n + 3 - k)^3}\int_0^\infty v^2 e^{-v}\,dv = \frac{1}{(6n + 3 - k)^3} \Gamma(3) = \frac{2}{(6n + 3 - k)^3}.$$ Thus $$F(k) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(6n + 3 - k)^3}$$ and $$I = F(1) + F(-1) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(6n + 2)^3} + 2\sum_{n=0}^\infty \frac{(-1)^n}{(6n + 4)^3} = \frac14 \sum_{n=0}^\infty \frac{(-1)^n}{(3n + 1)^3} + \frac14 \sum_{n=0}^\infty \frac{(-1)^n}{(3n + 2)^3}.$$ I have as yet no idea how to sum the series.

3. Jul 26, 2014

### Saitama

That's an interesting approach pasmith! It would be nice to see a method to solve that last sum.

4. Jul 26, 2014

### jostpuur

I figured out a way to compute this with complex integrals and residue theorem, but unfortunately the calculation appears to be four or five pages long, and it is extremely difficult to go it through without mistakes. My "first" answer to the question is

$$\frac{5\pi^3}{324\sqrt{3}}$$

I'm not sure if I would get the same result if I repeated the same calculation though...

5. Jul 29, 2014

### jostpuur

I was hoping that somebody who knows the answer would have confirmed my answer as right or wrong. Anyway, I'll start posting the details next, and I'll go over them again. Let's see where this goes... First we get over some preliminary stuff before the actual contour integrals.

$$1 + z^6 = 0 \quad\Longleftrightarrow\quad z = e^{\frac{\pi i}{6}+ \frac{n\pi i}{3}}\quad n\in\{0,1,\ldots,5\}$$

The zero $e^{\frac{\pi i}{6}}$ will turn out to be more important than the others. If we denote $g(z)=1+z^6$, then we have $g'(z)=6z^5$. The first order Taylor approximation around the important zero is

$$1 + z^6 = 6 e^{\frac{5\pi i }{6}}(z - e^{\frac{\pi i}{6}}) + O\big((z-e^{\frac{\pi i}{6}})^2\big)\quad\textrm{in}\quad z\to e^{\frac{\pi i}{6}}$$

which also implies

$$\frac{1}{1 + z^6} = \frac{e^{-\frac{5\pi i}{6}}}{6} \frac{1}{z - e^{\frac{\pi i}{6}}} + O(1)$$

We are going to need the trigonometric function values

$$\cos\Big(\frac{\pi}{6}\Big) = \frac{\sqrt{3}}{2}\quad\textrm{and}\quad \sin\Big(\frac{\pi}{6}\Big) = \frac{1}{2}$$

so I'll put these on display here to refresh the memory. If we need trigonometric function values for angles $\frac{\pi}{3}$, $\frac{2\pi}{3}$... and so on, we know how to reduce them to the mentioned ones.

We'll use a contour $\gamma = I_{1,R}\cup I_{2,R}\cup I_{3,R}$ where

$$I_{1,R} = [0,R]\subset \mathbb{R}$$
$$I_{2,R} = \big\{Re^{i\theta}\;\big|\; 0\leq \theta\leq \frac{\pi}{3}\big\}$$
$$I_{3,R} = \big\{re^{\frac{\pi i}{3}}\;\big|\; 0\leq r\leq R\big\}$$

seen as a path running counter clockwise. When $R>1$, out of the zeros of $1+z^6$ the $e^{\frac{\pi i}{6}}$ is only one circled around by the contour $\gamma$. If $f(z)$ is some complex analytic function, obviously we have

$$\int\limits_{I_{1,R}}f(z)dz = \int\limits_0^R f(x)dx$$

We also have

$$\int\limits_{I_{3,R}}f(z)dz = -e^{\frac{\pi i}{3}}\int\limits_0^R f(re^{\frac{\pi i}{3}})dr$$

On the right we have an ordinary integral over the set $[0,R]\subset\mathbb{R}$. On the left we have an integral over the set $I_{3,R}$ with the direction from right to left (and sloping downwards). The factor $-e^{\frac{\pi i}{3}}$ accounts for the integration direction. We will always want that

$$\int\limits_{I_{2,R}} f(z)dz \underset{R\to\infty}{\to} 0$$

In order to get this, it will be sufficient to check

$$\max_{0\leq \theta\leq \frac{\pi}{3}}|f(Re^{i\theta})| = o\Big(\frac{1}{R}\Big) \quad\textrm{in}\quad R\to\infty$$

In many case $f$ will approach zero faster than this.

We will also be interested in such $f$ which are not defined at the origin. Then we must consider the modifications

$$I_{1,\epsilon,R} = [\epsilon, R]\subset\mathbb{R}$$
$$I_{3,\epsilon,R} = \big\{re^{\frac{\pi i}{3}}\;\big|\; \epsilon \leq r\leq R\big\}$$
$$I_{4,\epsilon} = \big\{\epsilon e^{i\theta}\;\big|\; 0\leq \theta\leq \frac{\pi}{3}\big\}$$

$\gamma = I_{1,\epsilon, R}\cup I_{2,R}\cup I_{3,\epsilon,R}\cup I_{4,\epsilon}$. We will want that

$$\int\limits_{I_{4,\epsilon}} f(z)dz \underset{\epsilon\to 0}{\to} 0$$

For this it will be sufficient to prove that

$$\max_{0\leq \theta\leq \frac{\pi}{3}}|f(\epsilon e^{i\theta})| = o\Big(\frac{1}{\epsilon}\Big) \quad\textrm{in}\quad \epsilon\to 0$$

L'Hopital's rule can be used to prove

$$\lim_{R\to\infty} \frac{\log(R)}{R} = 0$$

$$\lim_{R\to\infty} \frac{\log^2(R)}{R} = 0$$

$$\lim_{\epsilon\to 0} \epsilon \log(\epsilon) = 0$$

$$\lim_{\epsilon\to 0} \epsilon \log^2(\epsilon) = 0$$

so we can use these when dealing with the integrals over the arcs.

Stuff continues in the following posts. Let's see if I can bring this to the end by referring to the mentioned facts.

6. Jul 29, 2014

### jostpuur

In this post I will prove the formula

$$\int\limits_0^{\infty} \frac{x^3}{1 + x^6}dx = \frac{\pi}{3\sqrt{3}}$$

We define a complex analytic function

$$f(z) = \frac{z^3}{1+z^6}$$

and use the contour mentioned in the preliminary post. We have

$$\int\limits_{I_{1,R}} f(z)dz = \int\limits_0^R \frac{x^3}{1+x^6}dx$$

and

$$\int\limits_{I_{3,R}}f(z)dz = e^{\frac{\pi i}{3}}\int\limits_0^R \frac{x^3}{1+x^6}dx$$

Here two minus signs have cancelled. First we have a factor $-e^{\frac{\pi i}{3}}$, but from the integrand we get $(re^{\frac{\pi i}{3}})^3 = -r^3$. On the arc $z\in I_{2,R}$ we have

$$|f(Re^{i\theta})| = \frac{R^3}{|1+R^6e^{6i\theta}|}\leq \frac{R^3}{R^6- 1} = \frac{1}{1 - \frac{1}{R^6}}\frac{1}{R^3} \leq \frac{2}{R^3}$$

with sufficienly large $R$, so the maxima of $|f(z)|$ on the arc is $o\big(\frac{1}{R}\big)$. In other words we have

$$\int\limits_{\gamma} f(z)dz = \Big(1 + e^{\frac{\pi i}{3}}\Big)\int\limits_0^R \frac{x^3}{1+x^6}dx + o(1)$$

In the limit $z\to e^{\frac{\pi i}{6}}$ the function has the form

$$f(z) = \frac{e^{-\frac{5\pi i}{6}}}{6} \frac{\big(e^{\frac{\pi i}{6}}\big)^3}{z - e^{\frac{\pi i}{6}}} + O(1)$$

so for all $R>1$, by the residue theorem, the integral also equals

$$2\pi i \frac{e^{-\frac{5\pi i}{6}}}{6} \big(e^{\frac{\pi i}{6}}\big)^3 = -\frac{\pi}{3}e^{-\frac{5\pi i}{6}}$$

Writing the two expressions for the integral as equal and taking the limit $R\to\infty$ gives

$$-\frac{\pi}{3} e^{-\frac{5\pi i}{6}} = \Big(1 + e^{\frac{\pi i}{3}}\Big)\int\limits_0^{\infty} \frac{x^3}{1+x^6}dx$$

The remaining task is the carefully solve the integral expression, and the result should be what I said it would be in the beginning of the post. A nice start would be to first multiply the both sides by $e^{-\frac{\pi i}{6}}$ so the we get a cosine on the right.

Last edited: Jul 29, 2014
7. Jul 29, 2014

### Saitama

@jostpuur: Your final answer is correct. I do not know much about contour integration but the solution isn't supposed to be very long and there is a very elementary solution to this problem.

8. Jul 29, 2014

### jostpuur

Well then this will be a lesson in how imagination and genius can be compensated with learned knowledge and hard work

This also means that if I manage to get my solution posted here, the problem will remain open for a simpler solution.

Thank's for the confimation anyway.

9. Jul 29, 2014

### jostpuur

In this post I will prove the formula

$$\int\limits_0^{\infty} \frac{x^3\log(x)}{1 + x^6}dx = \frac{\pi^2}{54}$$

We define a complex analytic function

$$f(z) = \frac{z^3\log(z)}{1 + z^6}$$

and choose the logarithm so that branch cut doesn't appear for the arguments $0\leq \theta\leq \frac{\pi}{3}$. By using the inequality proven earlier we get

$$|f(Re^{i\theta})| = \Big|\frac{z^3}{1+z^6}\Big| |\log(z)| \leq \frac{2(\log(R) + \frac{\pi}{3})}{R^3}$$

By using the result from L'Hopital's rule we get

$$\cdots = \frac{2}{R^2}\underbrace{\frac{\log(R)}{R}}_{\to 0} + \frac{2\pi}{3R^3} = o\Big(\frac{1}{R^2}\Big)$$

so this is $o\big(\frac{1}{R}\big)$ too, as needed. The function is not defined at origin, but we have

$$|f(\epsilon e^{i\theta})| \leq \frac{\epsilon^3 (\log(\epsilon) + \frac{\pi}{3})}{|1+\epsilon^6 e^{6i\theta}|} \leq 2\Big(\epsilon^2\cdot\underbrace{\epsilon\log(\epsilon)}_{\to 0} + \frac{\pi\epsilon^3}{3}\Big) = o(\epsilon^2)$$

So the integral will go to zero on both sets $I_{2,R}$ and $I_{4,\epsilon}$ in the limits $R\to \infty$ and $\epsilon\to 0$.

I just realized that the condition $o\big(\frac{1}{\epsilon}\big)$ was strange in the sense that if we only had this upper limit on the speed of divergence, we couldn't be sure that the integrals on $I_{1,\epsilon,R}$ and $I_{3,\epsilon,R}$ converged in the limit $\epsilon\to 0$. Anyway, now we know that the function $f$ converges to zero at the origin, and there is no problem with the convergence of integrals. Anyway, the function $f$ is still not analytic at the origin, so it makes sense to keep a little distance to it. So I believe it makes sense to use the set $I_{4,\epsilon}$.

We have

$$\int\limits_{I_{1,\epsilon,R}}f(z)dz = \int\limits_{\epsilon}^R \frac{x^3\log(x)}{1+x^6}dx$$

and

$$\int\limits_{I_{3,\epsilon,R}}f(z)dz = e^{\frac{\pi i}{3}}\int\limits_{\epsilon}^R \frac{x^3\big(\log(x) + \frac{\pi i}{3}\big)}{1+x^6}dx$$

and therefore

$$\int\limits_{\gamma} f(z)dz = \Big(1 + e^{\frac{\pi i}{3}}\Big)\int\limits_{\epsilon}^R \frac{x^3\log(x)}{1+x^6}dx+ \frac{\pi i}{3}e^{\frac{\pi i}{3}}\int\limits_{\epsilon}^R\frac{x^3}{1+x^6}dx + o(1)$$

On the other hand in the limit $z\to e^{\frac{\pi i}{6}}$ the function has the form

$$f(z) = \frac{e^{-\frac{5\pi i}{6}}}{6} \frac{(e^{\frac{\pi i}{6}})^3 \frac{\pi i}{6}}{z - e^{\frac{\pi i}{6}}} + O(1)$$

so by the residue theorem for all $0<\epsilon < 1 < R$ the integral has the value

$$2\pi i \frac{e^{-\frac{5\pi i}{6}}}{6} \big(e^{\frac{\pi i}{6}}\big)^3 \frac{\pi i}{6} = -\frac{\pi^2}{18} e^{-\frac{\pi i}{3}}$$

By writing the two expressions for the integral as equal, taking the limits $\epsilon\to 0$ and $R\to\infty$, and using the previous integral formula proven above, we get

$$-\frac{\pi^2}{18}e^{-\frac{\pi i}{3}} = \Big(1 + e^{\frac{\pi i}{3}}\Big)\int\limits_0^{\infty} \frac{x^3\log(x)}{1+x^6}dx + \frac{\pi i}{3}e^{\frac{\pi i}{3}}\frac{\pi}{3\sqrt{3}}$$

Then the remaining task is to carefully solve the unknown integral expression. The result should be what I said it would be in the beginning of this post.

10. Jul 29, 2014

### jostpuur

We define a complex analytic function

$$f(z) = \frac{z^3\log^2(z)}{1 + z^6}$$

and aim for the summit!

By using the L'Hopital's rule results mentioned above, we see that the integrals over the sets $I_{2,R}$ and $I_{4,\epsilon}$ converge to zero just like for the previous function too. The square of the logarithm has sufficiently similar asymptotics to the logarithm itself.

The formula

$$\int\limits_{I_{1,\epsilon,R}}f(z)dz = \int\limits_{\epsilon}^R\frac{x^3\log^2(x)}{1+x^6}dx$$

is obvious again. On the line $I_{3,\epsilon,R}$ the integral is not so obvious:

$$\int\limits_{I_{3,\epsilon,R}}f(z)dz = e^{\frac{\pi i}{3}}\int\limits_{\epsilon}^R \frac{x^3\big(\log^2(x) + \frac{2\pi i}{3}\log(x)- \frac{\pi^2}{9}\big)}{1 + x^6}dx$$

The integral over the whole contour is therefore

$$\int\limits_{\gamma}f(z)dz = \Big(1 + e^{\frac{\pi i}{3}}\Big)\int\limits_{\epsilon}^R \frac{x^3\log^2(x)}{1+x^6}dx + \frac{2\pi i}{3}e^{\frac{\pi i}{3}}\int\limits_{\epsilon}^R\frac{x^3\log(x)}{1+x^6}dx- \frac{\pi^2}{9}e^{\frac{\pi i}{3}}\int\limits_{\epsilon}^R \frac{x^3}{1+x^6}dx + o(1)$$

On the other hand in the limit $z\to e^{\frac{\pi i}{6}}$ the function has the form

$$f(z) = \frac{e^{-\frac{5\pi i}{6}}}{6} \frac{(e^{\frac{\pi i}{6}})^3 (\frac{\pi i}{6})^2}{z- e^{\frac{\pi i}{6}}} + O(1)$$

and by the residue theorem the value of the integral for $0<\epsilon < 1 < R$ is

$$2\pi i \frac{e^{-\frac{5\pi i}{6}}}{6} \big(e^{\frac{\pi i}{6}}\big)^3 \big(\frac{\pi i}{6}\big)^2 = -\frac{\pi^3}{4\cdot 27} e^{\frac{\pi i}{6}}$$

By writing the two expressions for the integral as equal, taking the limits $\epsilon\to 0$ and $R\to\infty$, and using the integral formulas proven above, we get

$$-\frac{\pi^3}{4\cdot 27}e^{\frac{\pi i}{6}} = \Big(1 + e^{\frac{\pi i}{3}}\Big)\int\limits_0^{\infty}\frac{x^3\log^2(x)}{1+x^6}dx + \frac{2\pi i}{3}e^{\frac{\pi i}{3}}\frac{\pi^2}{54} - \frac{\pi^2}{9}e^{\frac{\pi i}{3}}\frac{\pi}{3\sqrt{3}}$$

"So close, so far..."

We multiply both sides by $e^{-\frac{\pi i}{6}}$ and simplify:

$$-\frac{\pi^3}{4\cdot 27} = \sqrt{3}\int\limits_0^{\infty}\frac{x^3\log^2(x)}{1+x^6}dx + \frac{\pi^3}{3\cdot 27} e^{\frac{2\pi i}{3}} - \frac{\pi^3}{27\sqrt{3}}e^{\frac{\pi i}{6}}$$

We substitute

$$e^{\frac{2\pi i}{3}} = -\frac{1}{2} + \frac{i\sqrt{3}}{2}$$

and

$$e^{\frac{\pi i}{6}} = \frac{\sqrt{3}}{2}+ \frac{i}{2}$$

The imaginary terms cancel and we are left with

$$-\frac{\pi^3}{4\cdot 27} = \sqrt{3}\int\limits_0^{\infty} \frac{x^3\log^2(x)}{1+x^6}dx - \frac{\pi^3}{2\cdot 3^4} - \frac{\pi^3}{2\cdot 3^3}$$

We can solve the integral expression and it turns out to be

$$\int\limits_0^{\infty}\frac{x^3\log^2(x)}{1+x^6}dx = \frac{\pi^3}{2\cdot 3^3\sqrt{3}}\Big(\frac{1}{3} + 1 -\frac{1}{2}\Big) = \frac{5\pi^3}{2^2\cdot 3^4\sqrt{3}} = \frac{5\pi^3}{324\sqrt{3}}$$

It is difficult to go through all this without making mistakes, but little aid comes from the fact that mistakes usually result in an integral of some real valued function having an imaginary component. So eventually fixing the mistakes is not impossible.

My list consists of only Funktioteoria I-II by Olli Lehto. I didn't remember how these $2\pi i$ things went in the residue theorem, since I had not used complex integration for anything for years, and I had to check them from somewhere.

11. Jul 30, 2014

### chingel

I found a way to sum the series.

His series is

$$\frac{1}{4}(\frac{1}{1^3}+\frac{1}{2^3}-\frac{1}{4^3}-\frac{1}{5^3}+\frac{1}{7^3}+\frac{1}{8^3}-...)$$

Having seen the correct answer, since it had $\pi^3$ and the series has inverse third powers, it reminds of the method of using fourier series of polynomials to get various sums of inverse powers. So I tried something along those lines.

The fourier series from -1 to 1:

$$\frac{\pi^3}{12}(x-x^3)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^3} \sin(n \pi x)$$

Since the answer has a square root of 3 and every third element is missing, it is not too hard to put x=2/3 after a few tries.

$$\frac{\pi^3}{12}\frac{10}{27}=\frac{\sqrt{3}}{2}(\frac{1}{1^3}+\frac{1}{2^3}-\frac{1}{4^3}-\frac{1}{5^3}+\frac{1}{7^3}+\frac{1}{8^3}-...)$$

$$\frac{5\pi^3}{324\sqrt3}=\frac{1}{4}(\frac{1}{1^3}+\frac{1}{2^3}-\frac{1}{4^3}-\frac{1}{5^3}+\frac{1}{7^3}+\frac{1}{8^3}-...)$$

12. Jul 31, 2014

### pasmith

Without the hint of knowing the answer, it seems that the only way to express $$I = \frac14\sum_{n=0}^\infty \frac{(-1)^n}{(3n + 1)} + \frac14\sum_{n=0}^\infty \frac{(-1)^n}{(3n + 2)}$$ in a single summation of multiples of reciprocals of cubes of all positive integers is to write $$\frac{1}{1^3} + \frac{1}{2^3} - \frac{1}{4^3} - \frac{1}{5^3} + \dots = 1\cdot\frac{1}{1^3} + (-1)\cdot\frac{(-1)}{2^3} + 0\cdot\frac{1}{3^3} + 1\cdot\frac{(-1)}{4^3} + (-1)\cdot\frac{1}{5^3} + 0\cdot\frac{(-1)}{6^3} + \dots \\ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}a_n$$ where $a_n$ is periodic with period 3 and $a_0 = 0$, $a_1 = 1$, $a_2 = -1$. That being so, we can write $$a_n = A + B\sin\left(\frac{2n\pi}{3}\right) + C\cos\left(\frac{2n\pi}{3}\right)$$ and we find that $A = C = 0$ and $B = 2/\sqrt{3}$. Hence $$I = \frac{1}{2\sqrt{3}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} \sin\left(\frac{2n\pi}{3}\right)$$ and the right hand side is clearly a fourier series. I suppose one then has to know that the fourier series of $f : [-1:1] \to \mathbb{R} : x \mapsto x - x^3$ is $$\frac{12}{\pi^3} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} \sin(n\pi x).$$ Thus $$I = \frac{\pi^3}{24\sqrt{3}} f(\tfrac 23) = \frac{\pi^3}{24\sqrt{3}} \frac{10}{27} = \frac{5\pi^3}{324\sqrt{3}}.$$

13. Aug 4, 2014

Hello All.

It has been while since I posted here. I hope no one minds I give my two cents. Since I noticed Pranav frequents this site, I thought I would say howdy.

Anyway, something I often use when confronted with a sum like this is to use the digamma function.

Note the classic identity $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{a+n}=1/2\left(\psi(\frac{a+1}{2})-\psi(a/2)\right)$$

By differentiating this twice w.r.t 'a', one obtains:

$$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+a)^{3}}=1/16\left(\psi''(\frac{a+1}{2})-\psi''(a/2)\right)\tag{1}$$

But the sums in question are:

$$\frac{1}{108}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1/3)^{3}}+\frac{1}{108}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+2/3)^{3}}$$

By letting $$a=1/3$$ and $$a=2/3$$ in the formula in (1), we may obtain:

$$+\frac{1}{108}\left[\frac{39}{4}\zeta(3)+\frac{\pi^{3}}{12\sqrt{3}}+ \frac{\sqrt{3}\pi^{3}}{4}\right]$$
$$+\frac{1}{108} \left[-\frac{39}{4}\zeta(3)+\frac{\pi^{3}}{12\sqrt{3}}+ \frac{\sqrt{3}\pi^{3}}{4} \right]$$

Notice when we add these, the zeta terms cancel and we're left with:

$$\frac{5\pi^{3}\sqrt{3}}{972}$$ as required.

14. Aug 5, 2014

### Saitama

Hey fredoniahead! I am not sure if it is obvious, how did you find the values for the digammas?

Anyway, here's my solution.

Consider the integral:
$$I(n)=\int_0^{\infty} \frac{x^{n-1}}{1+x^{m}}\,dx$$
With the substitution, the integral is:
$$I(n)=\frac{1}{m}\int_0^1 y^{-n/m}(1-y)^{n/m-1}\,dy =B\left(\frac{n}{m},1-\frac{n}{m}\right)=\frac{\Gamma\left(\frac{n}{m}\right)\Gamma\left(1-\frac{n}{m}\right)}{\Gamma(1)}$$
$$\Rightarrow \int_0^{\infty} \frac{x^{n-1}}{1+x^{m}}\,dx=\frac{\pi}{m\sin\left(\frac{n}{m}\pi\right)}$$
Differentiate both the sides wrt $n$ twice:
$$\Rightarrow \int_0^{\infty} \frac{x^{n-1}\ln^2x}{1+x^{m}}\,dx=\frac{\pi^3}{m^3}\left(1+\cos^2\left(\frac{n\pi}{m}\right)\right)\csc^3\left(\frac{n}{m}\pi\right)$$
Substitute $n=4$ and $m=6$ to obtain the final result.

15. Aug 5, 2014