Caboodle of tricks

Let $J$ be the required integral. Consider the following integral with a parameter $a$ .
$I(a) = \int_0^{\infty} \frac{1}{1 + x^a} \text{ d}x$ $\displaystyle \Rightarrow I'(a) = \frac{\partial }{\partial a} \int_0^{\infty} \frac{1}{1 + x^a} \text{ d}x = - \int_0^{\infty} \frac{x^a \ln (x) }{(1 + x^a)^2} \text{ d}x = - J$ .

So evaluating $I(a)$ would be enough for evaluating $J$ . Infact $I(a)$ is a standard integral of the form $\displaystyle \int_0^{\infty} \frac{x^{m-1}}{1 + x^n} \text{ d}x$ with $m = 1$ and $n = a$ which evaluates to $\displaystyle \frac{\pi }{n} \csc \left(\frac{m \pi }{n} \right)$ .

So, $\displaystyle I(a) = \frac{\pi }{a} \csc \left(\frac{\pi }{a} \right)$ .

$\displaystyle \therefore I'(a) = \csc \left(\frac{\pi }{a} \right) \cot \left(\frac{\pi }{a} \right) \frac{ \pi ^2 }{a^3} - \csc \left(\frac{\pi }{a} \right)\frac{\pi }{a^2} = -J$ .

Substitute $a=4$ to get $\displaystyle J = \frac{\pi }{2^{7/2}} - \frac{\pi^2 }{2^{11/2}}$ .

Compare the coefficients and add them up to get the final answer as $\boxed{29}$ .

For the standard integral:

I will prove the general form of

$\int_0^\infty\frac{x^{m-1}}{1+x^n}\,dx=\frac{\pi}{n\sin\frac{m\pi}{n}}$

It can easily be proven by taking substitution $\displaystyle y=\frac{1}{1+x^n}$ and the integral becomes Beta function

$\frac{1}{n}\int_0^1 y^{\large 1-\frac{m}{n}-1}\ (1-y)^{\large \frac{m}{n}-1}\,dy=\frac{\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)}{n}=\frac{\pi}{n\sin\frac{m\pi}{n}}$

where the last part comes from Euler's reflection formula for the gamma function . Hence, by setting $m=1$ , we have

$\int_0^\infty\frac{1}{1+x^n}\,dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)$

$\iota \eta \tau \epsilon g \gamma \alpha \tau \iota \omicron \eta$

This solution uses contour integration.

We want to find the value of the integral ( $R > 1$ ) $J(m) = \int_{\gamma} \dfrac{1}{1 + z^m} dz$ where $\gamma$ is a sector in the complex plane, of radius $R$ and angle included from 0 to $\dfrac{2\pi}{m}$

Divide the sector into 3 parts ( $\gamma_1, \gamma_2, \gamma_3)$ that correspond to the arc and the two lines that bound it. $J(m) = \int_{\gamma_1} \dfrac{1}{1 + z^m} dz + \int_{\gamma_2} \dfrac{1}{1 + z^m} dz + \int_{\gamma_3} \dfrac{1}{1 + z^m} dz$ Observe that for the $\textbf{arc}$ (second part), the following argument holds. $\int_{\gamma_2} \dfrac{1}{1 + z^m} dz = \int_{\gamma_2} \dfrac{iR e^{i\phi}}{1 + (R e^{i\phi})^m} d\phi$ As $R \to \infty$ , the contribution of this integral $\int_{\gamma_2} \dfrac{iR e^{i\phi}}{1 + (R e^{i\phi})^m} d\phi \to 0$

For the third part, parametrize z as $z = xe^{2\pi i/ m}$ , by which the integral becomes $\int_{\gamma_3} \dfrac{1}{1 + z^m} dz = \int_{R}^{0} \dfrac{e^{2\pi i/m}}{1 + x^m(e^{2\pi})} dx$

Hence, $J(m) = \int_{\gamma} \dfrac{1}{1 + z^m} dz = \int_{0}^{\infty} \dfrac{1}{1 + x^m} dx + \int_{\infty}^{0} \dfrac{e^{2\pi i/m}}{1 + x^m} dx$ $J(m) = (1 - e^{2\pi i / m})\int_{0}^{\infty} \dfrac{1}{1 + x^m} dx$

By the method of residues, $\int_{\gamma} \dfrac{1}{1 + z^m} dz = \left. 2\pi i \times \text{Res}\left\{ \dfrac{1}{1 + z^m}\right\} \right|_{z = e^{\pi i/m}}$

Let $z_0 = e^{\pi i/m}$ . The residue is equal to $\lim_{z \to z_0} (z - z_0)\dfrac{1}{1 + z^m} = \lim_{z \to z_0} \dfrac{1}{m z^{m - 1}} = \dfrac{-z_0}{m}$ (Applying L'Hospital's Rule)

Hence, $\int_{0}^{\infty} \dfrac{1}{1 + x^m} dx = \dfrac{-2 \pi i}{m} \dfrac{e^{\pi i / m}}{1 - e^{2\pi i / m}}$

Rearranging, we get $\int_{0}^{\infty} \dfrac{1}{1 + x^m} dx = \dfrac{\pi}{m} \dfrac{1}{\sin\left(\dfrac{\pi}{m}\right)} = \dfrac{\pi}{m}\csc\left(\dfrac{\pi}{m}\right)$

Now, we know that $\int_{0}^{\infty} \dfrac{1}{1 + x^m} dx = I(m) \text{ (say)} = \dfrac{\pi}{m}\csc\left(\dfrac{\pi}{m}\right)$

The required integral is $-I'(4)$ which comes out to be $\displaystyle -I'(4) = \frac{\pi }{2^{7/2}} - \frac{\pi^2 }{2^{11/2}}$