Improper Integration of A Non-even Non-odd Function From $0$ to $\infty$ I am trying to calculate the integral:
$$\int_{0}^\infty  \frac{x^2dx}{1+x^7}$$
I used to face this type of integration with even integrand, but the function here is not even nor odd!
Is there a trick to do it?
 A: Start by substituting $u=x^3$, so $\frac{1}{3}du=x^2 dx$, and the integral becomes
$$ \frac{1}{3}\int_0^{\infty}\frac{1}{1+u^{\frac{7}{3}}}\, du,$$
and this integral can be evaluated using the well-known fact that 
$$\int_0^{\infty} \frac{1}{1+x^n}\, dx=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right).$$
See here for a proof. 
A: Let the integral $I$ be given by 
$$I=\int_0^{\infty}\frac{x^2}{1+x^7}dx$$
Then, enforcing the substitution $x\to \left(\frac{1-x}{x}\right)^{1/7}$ so that $x^2\to \left(\frac{1-x}{x}\right)^{1/7}$ and $dx \to -\frac17 x^{-2}\left(\frac{1-x}{x}\right)^{-6/7}$, $(1)$ becomes
$$\begin{align}
I&=\frac17 \int_0^1 x^{-3/7}\,(1-x)^{-4/7}\,dx\\\\
&=\frac17 B\left(\frac47,\frac37\right) \tag 2\\\\
&=\frac17\,\frac{\Gamma\left(\frac47\right)\Gamma\left(\frac37\right)}{\Gamma\left(\frac47+\frac37\right)} \tag 3\\\\
&=\frac17\,\Gamma\left(1-\frac37\right)\Gamma\left(\frac37\right) \tag 4\\\\
&=\frac17\,\frac{\pi}{\sin(3\pi/7)}\tag 5\\\\
&=\frac{\pi/7}{\sin(3\pi/7)}
\end{align}$$
Therefore, we have that the integral of interest $I$ is 
$$\bbox[5px,border:2px solid #C0A000]{I=\frac{\pi/7}{\sin(3\pi/7)}}$$

NOTES:
In arriving at $(2)$, we used the definition of the Beta Function, $B(x,y)=\int_0^1 t^{x-1}\,(1-t)^{y-1}\,dt$.
In going from $(2)$ to $(3)$, we used the Relationship between the Beta and Gamma Function, $B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$.
In going from $(4)$ to $(5)$ we used the Reflection Formula for the Gamma Function, namely, $\Gamma (z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$.
A: Another way to do it is to integrate
$$
f(z) = \frac{z^2\operatorname{Log}_n z}{1+z^7}
$$
around a keyhole contour. (Here $\operatorname{Log}_n z$ is the natural branch of the complex logarithm.)
See for example this for details. (There will be a lot of work doing this for this particular integrals though, with seven poles to take care of.)
