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.
On
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.
On
$$I=\int_{0}^{1}\frac{x^2}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2}{x^4+1}\,dx = \int_{0}^{1}\left(\frac{x^2}{x^4+1}+\frac{1}{x^4+1}\right)\,dx \tag{1}$$ 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: $$ \int_{0}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}} = \int_{0}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2}=\int_{0}^{+\infty}\frac{dx}{x^2+2}.\tag{5} $$
On
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 $$I=\frac{B\left(\frac{3}{4},\frac{1}{4}\right)}{4}=\color{red}{\frac{\pi}{2\sqrt{2}}}.$$
On
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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
$$ \bbx{\int_{0}^{\infty}{x^{2} \over x^{4} + 1}\,\dd x = {\root{2} \over 4}\,\pi} $$
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 $$f(x)=\frac{1}{2\sqrt{2}}\left(\frac{x}{x^2-\sqrt2x+1} -\frac{x}{x^2+\sqrt2x+1}\right).$$