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 $a$?

Solutions Collecting From Web of "Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$?"

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)} \\
\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}

Thus $\mathcal{I}(a) = \frac{\pi}{4}$ for all $a$. I do not see a need to require $a$ to be non-zero.

With a change of variable
Adding and dividing by two yields

$\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$


$\displaystyle x=\tan\theta$

$\displaystyle dx=\sec^2\theta d\theta$

$\displaystyle I=\int_0^{\pi/2}\frac{d\theta}{1+\tan^a \theta}$

$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a \theta d\theta}{\sin^a \theta + \cos^a \theta}$

$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a(\pi/2-\theta) d\theta}{\sin^{a}(\pi/2-\theta) + \cos^a (\pi/2-\theta)}$

$\displaystyle I=\int_0^{\pi/2}\frac{\sin^a\theta d\theta}{\sin^a\theta + \cos^a \theta}$


$\displaystyle 2I=\int_0^{\pi/2}d\theta$

$\displaystyle I=\pi/4$

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}

Let $\displaystyle J = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}$
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

Hence, $\frac{dI}{da}=0$. Hence, $I$ is independent of $a$.