42
$\begingroup$

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 $a$?

$\endgroup$
2
  • 4
    $\begingroup$ Duplicate answer, so to speak. $\endgroup$
    – Did
    Dec 2, 2011 at 22:36
  • 2
    $\begingroup$ It was already solved . Just use $\large x = \tan\left(\theta\right)$. $\endgroup$ Jan 27, 2014 at 23:31

4 Answers 4

55
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ I think there's a typo - you wanted to write $1+x^a$ instead of $1+x^2$ in the numerator. I hope you don't mind correcting. Very nice solution! $\endgroup$ Dec 2, 2011 at 14:13
  • $\begingroup$ can you explain the second step? thanks $\endgroup$
    – tomerg
    Dec 2, 2011 at 14:23
  • 2
    $\begingroup$ @tomerg $\mathrm{d}y = -\frac{1}{x^2}\mathrm{d}x$, $\frac{1}{1+y^2} = \frac{x^2}{1+x^2}$, and $\frac{1}{1+y^a}=\frac{x^a}{1+x^a}$. Also notice that $x \mapsto \frac{1}{x}$ reverts the orientation, i.e. changes boundaries $1$, $\infty$ into $1$, $0$ respectively. But the negative sign of Jacobian compensates that, thus overall +1 sign. $\endgroup$
    – Sasha
    Dec 2, 2011 at 15:35
  • 1
    $\begingroup$ +1 Nice argument. Is there a way to view this in terms of trigonometric identities involving arctangents? $\endgroup$ Dec 2, 2011 at 16:44
  • $\begingroup$ @MichaelHardy Thank you. Do you mean to perform substitution $u=\tan(x)$, so that $\mathcal{I}(a) = \int_0^{\pi/2} \left( 1 + \tan^a(u) \right)^{-1} \mathrm{d} u$ ? The idea of the proof will then remain much the same, split integration into $(0,\pi/4)$ and $(\pi/4, \pi/2)$ and use $u \mapsto \pi/2-u$ substitution for the second integral. Then due to $\tan(\pi/2-u) = \cot(u)$ the result would follow. $\endgroup$
    – Sasha
    Dec 2, 2011 at 18:01
34
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ I was just about to post this when I saw Sasha's answer appear so I deleted it. However, now that Sasha has probably capped for the day, I decided to undelete it. $\endgroup$
    – robjohn
    Dec 2, 2011 at 19:08
  • 4
    $\begingroup$ This is the simplest answer IMO. I upvoted this. :-) $\endgroup$
    – Srivatsan
    Dec 2, 2011 at 19:09
  • $\begingroup$ How does the numerator after the substitution become $x^a$? I would have expected $x^{2+a}$. $\endgroup$ Dec 3, 2011 at 12:39
  • 2
    $\begingroup$ @Henning: $$\begin{align}\int_\infty^0\frac{\mathrm{d}\frac{1}{x}}{\left(1+\left(\frac{1}{x}\right)^2\right)\left(1+\left(\frac{1}{x}\right)^a\right)}&=\int_\infty^0\frac{-\frac{1}{x^2}\;\mathrm{d}x}{(1+x^2)/x^2\;(1+x^a)/x^a}\\&=\int_0^\infty\frac{x^a\; \mathrm{d}x }{(1+x^2)(1+x^a)}\end{align}$$ I believe that is correct. $\endgroup$
    – robjohn
    Dec 3, 2011 at 15:24
  • $\begingroup$ Oh, of course. $ $ $\endgroup$ Dec 3, 2011 at 17:51
14
$\begingroup$

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

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,

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

$\displaystyle I=\pi/4$

$\endgroup$
2
13
$\begingroup$

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

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 $a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.