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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 $a$?

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

4 Answers 4

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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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  • $\begingroup$ 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! $\endgroup$ Commented Dec 2, 2011 at 14:13
  • $\begingroup$ can you explain the second step? thanks $\endgroup$
    – tomerg
    Commented Dec 2, 2011 at 14:23
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    $\begingroup$ @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
    Commented Dec 2, 2011 at 15:35
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    $\begingroup$ +1 Nice argument. Is there a way to view this in terms of trigonometric identities involving arctangents? $\endgroup$ Commented 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
    Commented Dec 2, 2011 at 18:01
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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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  • $\begingroup$ 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
    Commented Dec 2, 2011 at 19:08
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    $\begingroup$ This is the simplest answer IMO. I upvoted this. :-) $\endgroup$
    – Srivatsan
    Commented Dec 2, 2011 at 19:09
  • $\begingroup$ How does the numerator after the substitution become $x^a$? I would have expected $x^{2+a}$. $\endgroup$ Commented Dec 3, 2011 at 12:39
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    $\begingroup$ @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
    Commented Dec 3, 2011 at 15:24
  • $\begingroup$ Oh, of course. $ $ $\endgroup$ Commented Dec 3, 2011 at 17:51
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$\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$

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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  • $\begingroup$ Excellent! I think this should satisfy Michael Hardy's request. $\endgroup$
    – robjohn
    Commented Dec 3, 2011 at 15:13
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    $\begingroup$ Nice argument. +1 $\endgroup$ Commented Dec 6, 2011 at 4:47
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$$ \begin{align} I & = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x^a)}\\ \frac{dI}{da} & = -\int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2} \end{align} $$

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 $a$.

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