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

Question

Let $a$ be a non-zero real number.

Is it true that $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ is independent on $a$ ?

Any proof?

Answer 1

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.

Answer 2

$$ \begin{align} I & = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x^a)}\\ \frac{dI}{da} & = -\int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2} \end{align} $$

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 $a$.

Answer 3

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 $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)} &=\frac{1}{2}\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\ &=\frac{\pi}{4} \end{align} $$

Answer 4

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

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,

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

$\displaystyle I=\pi/4$