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 $a$?
Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$?
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2[It was already solved](http://math.stackexchange.com/a/605713/85343) . Just use $\large x = \tan\left(\theta\right)$. – 2014-01-27
4 Answers
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.
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0This is related to an integral challenge problem I made up: Show that $\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{\log\left(a^{2}+1\right)}{\left(1+x^{2}\right)\left(1+x^{a}\right)\left(1+a^{2}\right)}dxda=\pi^2 \log 2.$ Although, the problem is not so tricky now that you know the first step!! Without knowing that, it can be unclear how one should proceed. – 2012-08-31
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} $
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0Oh, of course. $ $ – 2011-12-03
$\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$
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,
$\displaystyle 2I=\int_0^{\pi/2}d\theta$
$\displaystyle I=\pi/4$
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1Nice argument. +1 – 2011-12-06
$ \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 $a$.