7
$\begingroup$

$$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$

I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=\tan\theta$, but I ended up with $\sin\theta\cos^3\theta$ as my integrand. Can someone explain how to compute this?

$\endgroup$
8
  • $\begingroup$ If you have $\sin\theta\cos^3\theta$ as your integrand, isn't it straightforward to substitute $u = \cos\theta$, and then $du = -\sin\theta d\theta$? $\endgroup$
    – Brian Tung
    Dec 8, 2015 at 2:49
  • $\begingroup$ Maybe, but I'm not sure that I substituted correctly. If $x^2 = \tan\theta$ I don't think I can day $dx = \sec^2\theta$. It would me $2xdx = \sec^2\theta$, which makes everything messy. $\endgroup$
    – Pareod
    Dec 8, 2015 at 2:52
  • $\begingroup$ Then try multiple subs. Allow that $y = x^2$, $dy = 2xdx$, then let $y=\tan(\phi)$, and then finally let $u = cos(\phi)$. $\endgroup$
    – Rellek
    Dec 8, 2015 at 2:54
  • $\begingroup$ Factor $x^4+1.$ $\endgroup$
    – Lucian
    Dec 8, 2015 at 2:55
  • 1
    $\begingroup$ My guess is that if the person asking wouldn't be attempting trig subs if they knew contour integration. $\endgroup$
    – Rellek
    Dec 8, 2015 at 2:58

13 Answers 13

7
$\begingroup$

Notice that: $$\begin{eqnarray*} \int_{-\infty}^{+\infty}\frac{x^2}{x^4+1}\,dx &=& 2\int_{0}^{+\infty}\frac{x^2}{x^4+1}\,dx\\ &=& 2\int_{0}^{1}\frac{x^2}{1+x^4}\,dx+2\int_{1}^{+\infty}\frac{x^2}{1+x^4}\,dx\\&=&2\int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx\\&=&2\left(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\ldots\right)\\&=&2\left(1+\frac{2}{4^2-1}-\frac{2}{8^2-1}+\frac{2}{12^2-1}-\ldots\right)\end{eqnarray*}$$ and, from the logarithmic derivative of the Weierstrass product for the sine and cosine function: $$ \sum_{k\geq 0}\frac{1}{(2k+1)^2-x^2}=\frac{\pi}{4x}\,\tan\left(\frac{\pi x}{2}\right),$$ $$ \sum_{k\geq 1}\frac{1}{k^2-x^2}=\frac{1-\pi x\cot(\pi x)}{2x^2},$$ so by taking limits as $x\to\frac{1}{2}$ we get: $$ \int_{-\infty}^{+\infty}\frac{x^2}{x^4+1}\,dx = \color{red}{\frac{\pi}{\sqrt{2}}}.$$

$\endgroup$
5
$\begingroup$

$$\dfrac{2x^2}{x^4+1}=\dfrac{x^2+1}{x^4+1}+\dfrac{x^2-1}{x^4+1}$$

$$=\dfrac{1+1/x^2}{(x-1/x)^2+2}+\dfrac{1-1/x^2}{(x+1/x)^2-2}$$

For the first integral, set $x-1/x=u$ and can you guess the substitution for the second?

$\endgroup$
4
$\begingroup$

Notice $$\int_{-\infty}^\infty \frac{x^2}{x^4+1} dx = 2 \int_0^\infty \frac{x^2}{x^4+1} dx = 2 \int_0^\infty \frac{dx}{x^2+x^{-2}} = 2 \int_0^\infty \frac{dx}{(x-x^{-1})^2+2}\\ = 2 \left(\color{red}{\int_0^1} + \color{blue}{\int_1^\infty}\right) \frac{dx}{(x-x^{-1})^2+2} $$ Change variable from $x$ to $\frac1x$ for that part of integral on $(0,1)$, this becomes $$2\int_1^\infty \frac{1}{(x-x^{-1})^2+2}(\color{blue}{1}+\color{red}{x^{-2}})dx = 2\int_1^\infty \frac{d(x-x^{-1})}{(x-x^{-1})^2+2} $$ Change variable once more to $y = x-x^{-1}$ and then to $y = \sqrt{2}z$, we get $$\int_{-\infty}^\infty \frac{x^2}{x^4+1} dx = 2 \int_0^\infty \frac{dy}{y^2+2} = \int_{-\infty}^\infty \frac{dy}{y^2+2} = \frac{1}{\sqrt{2}}\int_{-\infty}^{\infty} \frac{dz}{1+z^2} = \frac{\pi}{\sqrt{2}}$$

$\endgroup$
3
$\begingroup$

You have $x^4+1=(x^2+x\sqrt 2+1)(x^2-x\sqrt 2+1)$ Now you can do partial fractions. After that, a trig substitution will be your friend. Trig substitutions work well with quadratics, not so well with higher powers.

$\endgroup$
2
  • $\begingroup$ Would the numerators be of the form $Ax^2 + Bx +C$? $\endgroup$
    – Pareod
    Dec 8, 2015 at 3:09
  • 1
    $\begingroup$ Each numerator would be of the form Ax + B $\endgroup$
    – Kaynex
    Dec 8, 2015 at 3:18
3
$\begingroup$

The most straightforward method is to use the residue theorem.

Let $f(z)=\frac{z^2}{z^4+1}$. This function has exactly two poles in the upper half plane of $\mathbb{C}$: $$z_1=e^{i\frac{\pi}{4}}$$ $$z_2=e^{i\frac{3\pi}{4}}$$

Calculate residues at that points:

$$\operatorname{res}_f(z_1)=\lim_{z\rightarrow z_1} \frac{z^2}{(z-e^{-i\frac{\pi}{4}})(z-e^{-i\frac{3\pi}{4}})(z-e^{i\frac{3\pi}{4}})}=\frac{1}{4}e^{-i \frac{\pi}{4}}$$

$$\operatorname{res}_f(z_2)=\lim_{z\rightarrow z_2} \frac{z^2}{(z-e^{-i\frac{\pi}{4}})(z-e^{-i\frac{3\pi}{4}})(z-e^{i\frac{\pi}{4}})}=\frac{1}{4}e^{-i \frac{3\pi}{4}}$$

Now we will use the residue theorem.

enter image description here

Let $R>0$ be big enough so the upper semicircle contains $z_1$ and $z_2$. From the residue theorem the integral of $f$ over the above contour of integration is equal to $$2\pi i (\operatorname{res}_f(z_1)+\operatorname{res}_f(z_2))$$ It is easy to verify that when $R\rightarrow \infty$, then the integral over the semicircle part of that contour vanishes, so the contour integral converges to our initial integral $\int_{-\infty}^\infty \frac{x^2}{x^4+1} dx$. Thus

$$\int_{-\infty}^\infty \frac{x^2}{x^4+1} dx=2\pi i (\operatorname{res}_f(z_1)+\operatorname{res}_f(z_2))=\frac{\pi}{\sqrt{2}}$$

$\endgroup$
1
  • $\begingroup$ Awesome answer, but I don't know if the OP knows contour integration yet. But in reality, this is the simplest to do answer. I like it. But of course the limits had to do with it's simplicity. $\endgroup$ Oct 26, 2019 at 14:15
2
$\begingroup$

your substitution is perfectly ok, if you convert your integral to :$$I=2\int_{0}^{\infty}\frac{x^2 \mathrm{d}x}{x^4+1}$$ now by $x^2=\tan\theta$ we get

$$I=\int_{0}^{\frac{\pi}{2}}\sqrt{\tan\theta}\mathrm{d}\theta \tag{1}$$

now to evaluate $I$ consider its complement $$I=\int_{0}^{\frac{\pi}{2}}\sqrt{\cot\theta}\mathrm{d}\theta \tag{2}$$ adding both

$$2I=\int_{0}^{\frac{\pi}{2}}\left(\sqrt{\tan\theta}+\sqrt{\cot\theta}\right)\mathrm{d}\theta=2\int_{0}^{\frac{\pi}{4}}\left(\sqrt{\tan\theta}+\sqrt{\cot\theta}\right)\mathrm{d}\theta=2\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin\theta+\cos\theta}{\sqrt{\sin\theta \cos\theta}}\right)$$ and use $\sin\theta \cos\theta=\frac{1}{\sqrt{2}}\sqrt{1-(\sin\theta-\cos\theta)^2}$ and use again the substitution $\sin\theta-\cos\theta=t$.

Hope you can take it from here

$\endgroup$
7
  • $\begingroup$ I'm having trouble understanding how you got $(1)$. Wouldn't the sub cause $2\int_{-\infty}^\infty \frac {\tan\theta} {\tan^2\theta + 1} d\theta$? $\endgroup$
    – Pareod
    Dec 8, 2015 at 3:24
  • $\begingroup$ yes but what is $dx$, it is $\frac{sec^2\theta d\theta}{2x}=\frac{ sec^2\theta d\theta}{2\sqrt{tan\theta}}$. so you missed there.. $\endgroup$ Dec 8, 2015 at 3:28
  • $\begingroup$ So would I also have to find $\int_0^{\pi/2} (\sqrt{\tan\theta}-\sqrt{\cot\theta})d\theta$ and then add both results together? $\endgroup$
    – Pareod
    Dec 8, 2015 at 4:03
  • $\begingroup$ for definite integral with the above limits its not required. if you keenly observe the integral you commented above is zero, isn't it? $\endgroup$ Dec 8, 2015 at 4:05
  • $\begingroup$ Okay, so $\int_0^{\pi/2} \sqrt{\tan\theta} d\theta = \int_0^{\pi/2} \sqrt{\tan\theta} + \sqrt{cot\theta} d\theta$? All I need to do is get rid of my subs and take the limit at my bound approaches infinity? $\endgroup$
    – Pareod
    Dec 8, 2015 at 4:14
2
$\begingroup$

Use the results from Jack's to establish the first inequality below, and for the third use $x \to \tan x$. The rest is standard stuff. \begin{align} I&=2\int_{0}^{\infty}\frac{x^2}{x^4+1}dx\\ &=2\int_{0}^{1}\frac{1+x^2 }{x^4+1}dx\\ &=2\int_0^{\pi/4}\frac{\left(\tan ^2x+1\right) \sec ^2x}{\tan ^4x+1}dx\\ &=2\int_0^{\pi/4}\frac{1}{\cos ^4x+\sin^4x}dx\\ &=2\int_0^{\pi/4}\frac{1}{(\cos ^2x-\sin^2x)^2+2\sin^2x\cos^2x}dx\\ &=2\int_0^{\pi/4}\frac{1}{\cos^2 2x+\frac12\sin^2 2x}dx\\ &=2\int_0^{\pi/4}\frac{\sec^2 2x}{1+\frac12\tan^2 2x}dx\\ &=2\int_0^{\pi/4}\frac{\sec^2 2x}{1+\frac12\tan^2 2x}dx\\ &=2 \times \frac{1}{\sqrt 2}\arctan\Big(\frac{1}{\sqrt2} \tan 2x\Big) \Big|_0^{\pi/4}\\ &=\frac{\pi}{\sqrt 2} \end{align}

$\endgroup$
2
$\begingroup$

In fact, using the transform $t=x^4$ and $t+1=\frac1{1-u}$ \begin{eqnarray} \int_{-\infty}^\infty\frac{x^2}{x^4+1}dx&=& 2\int_0^\infty\frac{x^2}{x^4+1}dx\\ &=&\frac12\int_0^\infty\frac{1}{\sqrt[4]{t^3}(t+1)}dt\\ &=&\frac12\int_0^1(1-u)^{-\frac14}u^{-\frac34}du\\ &=&\frac12B(\frac14,\frac34)\\ &=&\frac12\frac{\pi}{\sin\frac{3}{4}\pi} \end{eqnarray}

$\endgroup$
0
1
$\begingroup$

Notice that: $$x^4 + 1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2 + 1)^2 - 2x^2$$

$\endgroup$
4
  • $\begingroup$ So this would be of the form $\sqrt {y^2 + a^2}$ where $y = (x^2 + 1)^2$ and $a = \sqrt 2x$? $\endgroup$
    – Pareod
    Dec 8, 2015 at 3:07
  • $\begingroup$ So what would follow up. $\endgroup$ Dec 8, 2015 at 3:07
  • $\begingroup$ Did you try for $\frac{1}{a}tan^{-1}\frac{x}{a}$ $\endgroup$ Dec 8, 2015 at 3:09
  • $\begingroup$ Realized now I made a mistake in my work. So, I'm not sure where to go from here. You can treat it as a difference of squares like Ross Millikan did above, and that seems to be a good solution. $\endgroup$
    – Kaynex
    Dec 8, 2015 at 3:17
1
$\begingroup$

Note\begin{align*} \int_{-\infty}^{\infty} \frac{x^2}{x^4+1} dx &\overset{x\to \frac1x}=\int_{-\infty}^{\infty} \frac{1}{x^4+1}dx =\int_{0}^{\infty} \frac{x^2+1}{x^4+1}dx =\int_{0}^{\infty} \frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}dx \\ &= \int_{0}^{\infty} \frac{d(x-\frac{1}{x} )}{\left(x-\frac{1}{x}\right)^2+2} =\frac{1}{\sqrt{2}} \tan^{-1}{\frac{x-\frac{1}{x}}{\sqrt{2}}} \bigg \rvert_{0}^{\infty} = {\frac{\pi}{\sqrt{2}}} \end{align*}

$\endgroup$
1
  • $\begingroup$ Fancy! ${}{}{}{}{}$ $\endgroup$
    – mjw
    Sep 7, 2020 at 19:00
0
$\begingroup$

\begin{align*} I=\int_{-\infty}^{\infty} \frac{x^2}{x^4+1} \; \mathrm{d}x &= \int_{-\infty}^{\infty} \frac{1}{\left(x-\frac{1}{x}\right)^2+2} \; \mathrm{d}x\\ &=\int_{-\infty}^{\infty} \frac{1}{x^2+2} \; \mathrm{d}x \tag{1}\\ &=\frac{1}{\sqrt{2}} \arctan{\left(\frac{x}{\sqrt{2}}\right)} \bigg \rvert_{-\infty}^{\infty} \\ &= \boxed{\frac{\pi}{\sqrt{2}}} \end{align*} Where Glasser's master theorem was used to arrive to (1).

$\endgroup$
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over x^{4} + 1}\,\dd x} = 2\int_{0}^{\infty}{x^{2} \over x^{4} + 1}\,\dd x \,\,\,\stackrel{x^{\large 4}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{x^{-1/4} \over 1 + x}\,\dd x \label{1}\tag{1} \end{align} However, $\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty}\pars{-x}^{k} = \sum_{k = 0}^{\infty}\color{blue}{\Gamma\pars{k + 1}}{\pars{-x}^{k} \over k!}}$.

With Ramanujan's Master Theorem, (\ref{1}) becomes \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over x^{4} + 1}\,\dd x} = {1 \over 2}\int_{0}^{\infty}{x^{\color{red}{3/4} - 1} \over x^{4} + 1}\,\dd x = {1 \over 2}\,\Gamma\pars{\color{red}{3 \over 4}} \color{blue}{\Gamma\pars{-\,\color{red}{3 \over 4} + 1}} \\[5mm] = &\ {1 \over 2}\,{\pi \over \sin\pars{\pi/4}} = \bbx{{\root{2} \over 2}\,\pi} \\ & \end{align}

$\endgroup$
0
$\begingroup$

Via contour integration $\oint_C \frac{z^{\alpha-1}}{{1+z}} dz$ it was shown that

$$\int_0^\infty \frac{y^{\alpha-1}}{1+y} dy = \frac{\pi}{\sin \pi \alpha}.$$

Let $y=x^4$ so that $dy = 4x^3 dx.$

$$\int_{-\infty}^\infty \frac{x^2}{1+x^4} dx =\frac{1}{4} \int_{-\infty}^\infty \frac{y^{-1/4}}{1+y} dy = \frac{1}{2} \int_0^\infty \frac{y^{-1/4}}{1+y}dy=\frac{1}{2} \frac{\pi}{\sin \frac{3\pi}{4}}=\frac{\pi}{\sqrt{2}}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .