# Existence/Limit of Improper Integral

I got the improper integral: $$\int_0^\infty \frac{x^2}{x^4+1} \, dx$$

On one hand one could of course just evaluate the integral and then apply limits. However, it's not always practical to find an antiderivative. I got a hint that one can trace back improper integrals to sequences and thus just calculate the limit of a sequence which is way easier, could someone explain to me how this works or link me to some theorems? Thanks.

• This looks like a standard candidate for evaluation by contour integration. – Travis Willse May 30 '17 at 14:44
• Are you actually trying to evaluate it or just trying to show that it has a finite value? – tilper May 30 '17 at 14:44
• I just need to show it has a finite value, and i haven't had contour integration sorr :/ – johnka May 30 '17 at 14:50

The value of this integral is finite because $f(x):={ \frac { x^ 2 }{ x^ 4+1 } }$ is continuous in $\mathbb{R}$ and $f(x)\sim 1/x^2$ as $x$ goes to $\pm\infty$.

It can be evaluated by using Residue Theorem: $$\int _{ 0 }^{ \infty }f(x)dx= \frac{1}{2}\int _{ -\infty }^{ \infty }f(x)dx= \pi i(\mbox{Res}(f,e^{i\pi/4})+\mbox{Res}(f,e^{i3\pi/4}))\\= \pi i\left(\frac{1}{4e^{i\pi/4}}+\frac{1}{4ie^{i\pi/4}}\right)= \frac{\pi}{2\sqrt{2}}.$$

Alternatively, you can integrate $f$ by noting that $$x^4+1 =(x^2+1)^2 - (\sqrt2x)^2 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$$ Then, by using partial fraction decomposition, we get $$f(x)=\frac{1}{2\sqrt{2}}\left(\frac{x}{x^2-\sqrt2x+1} -\frac{x}{x^2+\sqrt2x+1}\right).$$

• Sorry I'm not familiar with complexe Analysis yet – johnka May 30 '17 at 14:54
• Then try the second approach – Robert Z May 30 '17 at 14:58
• I have already tried the second approch before but its fairly long, and i think i formulated my question wrong i only need to show the existence :) – johnka May 30 '17 at 15:22
• @johnka For the convergence my first two lines suffice. – Robert Z May 30 '17 at 15:29

Based on the overall nature of your post (and the comment you left that I saw as I finished this), it sounds like you're trying to show this integral actually has a finite value and you're not necessarily trying to find what that value is.

Since the integrand is $\dfrac{x^2}{x^4+1}$, then we know that as $x$ gets really super huge (i.e., as $x \to +\infty$), the integrand "behaves like" $\dfrac{x^2}{x^4} = \dfrac1{x^2}$. And we know that $\displaystyle \int_1^{+\infty} \dfrac1{x^2} \, dx$ has a finite value. Actually we can calculate it directly, and it's $1$. (This does not mean the original integral is also $1$. We can't conclude anything about the original integral yet!)

So, because our integrand $\dfrac{x^2}{x^4+1}$ "behaves like" an integrand that has a finite integral over $[1,+\infty)$, we expect that our integral of $\dfrac{x^2}{x^4+1}$ over $[0,+\infty)$ is also finite.

To prove it, we can just use the direct comparison test. On $[1, +\infty)$, we know that $x^4 \le x^4 + 1$. Divide both sides by $x^2(x^4+1)$ to conclude that $\dfrac{x^2}{x^4+1} \le \dfrac1{x^2}$. Since $\displaystyle \int_1^{+\infty} \dfrac1{x^2} \, dx$ is finite, then so is $\displaystyle \int_0^{+\infty} \dfrac{x^2}{x^4+1} \, dx$.

Note 1: Technically the above directly implies that $\displaystyle \int_1^{+\infty} \dfrac{x^2}{x^4+1} \, dx$ is finite, but $\displaystyle \int_0^1 \dfrac{x^2}{x^4+1} \, dx$ is clearly finite and so we conclude $\displaystyle \int_0^1 \dfrac{x^2}{x^4+1} \, dx + \displaystyle \int_1^{+\infty} \dfrac{x^2}{x^4+1} \, dx = \displaystyle \int_0^{+\infty} \dfrac{x^2}{x^4+1} \, dx$ is finite.

Note 2: For the comparison test for integrals to work, you also need to note that $\dfrac{x^2}{x^4+1} \ge 0$ on $[0,+\infty)$ (technically you just need on $[1,+\infty)$). This is obviously true, but to be thorough it really should be at least mentioned.

• Thanks, good answer. Could you argue with f is continious on the Intervall (0,1) and therefor you can integrate it Thus watching only the intervall (1,INF) because we know that the statement is true in the other fraction? – johnka May 30 '17 at 15:20
• @johnka, yes, since $\dfrac{x^2}{x^4+1}$ is continuous on $[0,1]$, it has a maximum value on $[0,1]$. Say this maximum value is $M$. Then $\displaystyle \int_0^1 \dfrac{x^2}{x^4+1} \, dx \le M(1-0) = M$, and so all we really need to worry about is whether or not $\displaystyle \int_1^{+\infty} \dfrac{x^2}{x^4+1} \, dx$ is finite, and it is by the integral comparison test. – tilper May 30 '17 at 15:31

$$I=\int_{0}^{1}\frac{x^2}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2}{x^4+1}\,dx = \int_{0}^{1}\left(\frac{x^2}{x^4+1}+\frac{1}{x^4+1}\right)\,dx \tag{1}$$ leads to: $$I = \int_{0}^{1}\frac{1+x^2-x^4-x^6}{1-x^8}\,dx = \sum_{k\geq 0}\left(\frac{1}{8k+1}+\frac{1}{8k+3}-\frac{1}{8k+5}-\frac{1}{8k+7}\right)\tag{2}$$ but due to the digamma reflection formula we have: $$\sum_{k\geq 0}\left(\frac{1}{8k+1}-\frac{1}{8k+7}\right)=\frac{\pi}{8}\cot\frac{\pi}{8}, \quad \sum_{k\geq 0}\left(\frac{1}{8k+3}-\frac{1}{8k+5}\right)=\frac{\pi}{8}\cot\frac{3\pi}{8}\tag{3}$$ hence it follows that: $$I = \frac{\pi}{8}\left[\cot\frac{\pi}{8}+\cot\frac{3\pi}{8}\right]=\color{red}{\frac{\pi}{2\sqrt{2}}}.\tag{4}$$ The same can be achieved by partial fraction decomposition, since $x^4+1=\Phi_8(x)=(x^2-x\sqrt{2}+1)(x^2+x\sqrt{2}+1)$.

Still another way is to apply Glasser's master theorem: $$\int_{0}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}} = \int_{0}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2}=\int_{0}^{+\infty}\frac{dx}{x^2+2}.\tag{5}$$

Note that $$I=\int_{0}^{\infty}\frac{x^{2}}{x^{4}+1}dx\stackrel{x^{4}=u}{=}\frac{1}{4}\int_{0}^{\infty}\frac{u^{-1/4}}{u+1}du$$ and now we can see that this is one of the form of the Beta function. So $$I=\frac{B\left(\frac{3}{4},\frac{1}{4}\right)}{4}=\color{red}{\frac{\pi}{2\sqrt{2}}}.$$


Lets $\ds{\mc{C}_{R}}$ a quarter circumference in the first quadrant with radius $\ds{R > 1}$ and vertex at the origin. Such contour 'encloses' a single pole $\ds{p \equiv \expo{\ic\pi/4}}$.

\begin{align} \Re\oint_{\mc{C}_{R}}{z^{2} \over z^{4} + 1}\,\dd z & = \Re\pars{2\pi\ic\,\lim_{z \to p}{\bracks{z - p}z^{2} \over z^{4} - 1}} = -2\pi\,\Im\pars{3p^{2} - 2p^{2} \over 4p^{3}} \\[5mm] & = -\,{1 \over 2}\,\pi\,\Im\pars{1 \over p} = -\,{1 \over 2}\,\pi\,\Im\pars{\expo{-\ic\pi/4}} = {\root{2} \over 4}\,\pi\label{1}\tag{1} \\[1cm] \Re\oint_{\mc{C}_{R}}{z^{2} \over z^{4} + 1}\,\dd z & = \int_{0}^{R}{x^{2} \over x^{4} + 1}\,\dd x \\[3mm] & +\ \underbrace{\Re\pars{\int_{0}^{\pi/2}{R^{2}\expo{2\ic\theta} \over R^{4}\expo{4\ic\theta} + 1}\,R\expo{\ic\theta}\ic\,\dd\theta}} _{\ds{\to\ 0\ \mrm{as}\ R\ \to\ \infty}}\ +\ \underbrace{% \Re\pars{\int_{R}^{0}{-y^{2} \over y^{4} + 1}\,\ic\,\dd y}}_{\ds{=\ 0}} \label{2}\tag{2} \end{align}

\eqref{1} and \eqref{2} lead, in the $\ds{R \to \infty}$ limit, to

$$\bbx{\int_{0}^{\infty}{x^{2} \over x^{4} + 1}\,\dd x = {\root{2} \over 4}\,\pi}$$

ANOTHER METHOD: \begin{align} \int_{0}^{\infty}{x^{2} \over x^{4} + 1}\,\dd x & = \int_{0}^{\infty}{\dd x \over x^{2} + 1/x^{2}} = \int_{0}^{\infty}{\dd x \over \pars{x - 1/x}^{2} + 2} \\[5mm] & = {1 \over 2}\bracks{% \int_{0}^{\infty}{\dd x \over \pars{x - 1/x}^{2} + 2} + \int_{\infty}^{0}{1 \over \pars{1/x - x}^{2} + 2}\pars{-\,{\dd x \over x^{2}}}} \\[5mm] & = {1 \over 2} \int_{0}^{\infty}{1 + 1/x^{2} \over \pars{x - 1/x}^{2} + 2} \,\dd x \,\,\,\stackrel{t\ =\ x - 1/x}{=}\,\,\, {1 \over 2}\int_{-\infty}^{\infty}{\dd t \over t^{2} + 2} \\[5mm] & = {1 \over \root{2}}\int_{0}^{\infty}{\dd t \over t^{2} + 1} = {\root{2} \over 2}\,{\pi \over 2} = \bbx{{\root{2} \over 4}\,\pi} \end{align}