Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$? Let $a$ be a non-zero real number. Is it true that the value of $$\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$$  is independent on $a$?
 A: Let $\mathcal{I}(a)$ denote the integral. Then
$$ \begin{eqnarray}
  \mathcal{I}(a) &=& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_1^\infty \frac{\mathrm{d} y}{(1+y^2)(1+y^a)} \\
   &\stackrel{y=1/x}{=}&
     \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + 
     \int_0^1 \frac{x^a \mathrm{d} x}{(1+x^2)(1+x^a)}   \\
   &=& \int_0^1 \frac{1+x^a}{(1+x^2)(1+x^a)} \mathrm{d} x = 
\int_0^1 \frac{1}{1+x^2} \mathrm{d} x = \frac{\pi}{4}
\end{eqnarray}
$$
Thus $\mathcal{I}(a) = \frac{\pi}{4}$ for all $a$. I do not see a need to require $a$ to be non-zero.
A: With a change of variable
$$
\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)}\overset{x\to1/x}{=}\int_0^\infty\frac{x^a\mathrm{d}x}{(1+x^2)(1+x^a)}
$$
Adding and dividing by two yields
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)}
&=\frac{1}{2}\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\
&=\frac{\pi}{4}
\end{align}
$$
A: $\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$
Substitution:
$\displaystyle x=\tan\theta$
$\displaystyle dx=\sec^2\theta d\theta$
$\displaystyle I=\int_0^{\pi/2}\frac{d\theta}{1+\tan^a \theta}$
$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a \theta d\theta}{\sin^a \theta + \cos^a \theta}$ 
$\displaystyle  I=\int_0^{\pi/2}\frac{\cos^a(\pi/2-\theta) d\theta}{\sin^{a}(\pi/2-\theta) + \cos^a (\pi/2-\theta)}$
$\displaystyle I=\int_0^{\pi/2}\frac{\sin^a\theta d\theta}{\sin^a\theta + \cos^a \theta}$ 
Therefore,
$\displaystyle 2I=\int_0^{\pi/2}d\theta$
$\displaystyle I=\pi/4$
A: $$
\begin{align}
I & = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x^a)}\\
\frac{dI}{da} & = -\int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}
\end{align}
$$
Let $\displaystyle J = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}$
$$
\begin{align}
J & = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}\\
& \stackrel{x=1/y}{=} \int_{0}^{\infty} \frac{1/y^a \log(1/y) d(1/y)}{(1+(1/y)^2)(1+(1/y)^a)^2}\\
& = \int_{\infty}^{0} \frac{y^a \log(y) dy}{(1+y^2)(1+y^a)^2}\\
& = -J
\end{align}
$$
Hence, $\frac{dI}{da}=0$. Hence, $I$ is independent of $a$.
