How to calculate $ \int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4} $?

Question

I would like to calculate $$\int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4}$$ by means of the Residue Theorem. This is what I tried so far: We can define a path $\alpha$ that consists of half a half-circle part ($\alpha_r$) and a path connecting the first and last point of that half circle (with radius $r$) so that we have $$ \int_{-r}^{r} f(x) dx + \int_{\alpha_r} f(z) dz = \int_{\alpha} f(z) dz = 2 \pi i \sum_{v = 1}^{k} \text{Res}(f;a_v) $$ where $a_v$ are zeros of the function $\frac{x^2 \log(x) }{1+x^4}$.

If we know $$\lim_{r \to \infty} \int_{\alpha_r} f(z) dz = 0 \tag{*} $$ then we know that $$\lim_{r \to \infty} \int_{-r}^{r} f(x) dx = \int_{-\infty}^{\infty} f(x) dx = 2 \pi i \sum_{v=1}^{k} \text{Res}(f;a_v) $$ and it becomes 'easy'.

Q: How do we know (*) is true?

Answer 1

It's a bit more tricky that what you describe, but the general idea is correct. Instead of integrating from $0$ to $\infty$, one can integrate from $-\infty$ to $+\infty$ slightly above the real axis. Because of the logarithm, the integral from $-\infty$ to $0$ will give a possibly non-zero imaginary part, but the real part will be an even function of $x$. So we can write: \begin{align} \int_0^{\infty}\frac{x^2\ln x}{1+x^4}dx&=\frac12\mathrm{Re}\,\int_{-\infty+i0}^{\infty+i0} \frac{x^2\ln x}{1+x^4}dx=\\&=\pi\cdot \mathrm{Re}\left[ i\left(\mathrm{res}_{x=e^{i\pi/4}}\frac{x^2\ln x}{1+x^4}+\mathrm{res}_{x=e^{3i\pi/4}}\frac{x^2\ln x}{1+x^4}\right)\right]=\\ &=\pi\cdot \mathrm{Re}\left[ i\left(\frac{\pi e^{i\pi/4}}{16}- \frac{3\pi e^{3i\pi/4}}{16}\right)\right]=\\ &=\pi\cdot\mathrm{Re}\frac{(1+2i)\pi}{8\sqrt{2}}=\frac{\pi^2}{8\sqrt{2}}. \end{align}

Now as far as I understand the question was about how can one justify the vanishing of the integral over the half-circle $C$ which in its turn justifies the application of residue theorem. Parameterizing that circle as $x=Re^{i\varphi}$, $\varphi\in(0,\pi)$, we see that \begin{align} \int_C \frac{x^2\ln x}{1+x^4}dx=\int_0^{\pi}\frac{iR^3e^{3i\varphi}\left(i\varphi+\ln R\right)}{1+R^4e^{4i\varphi}}d\varphi=O\left(\frac{\ln R}{R}\right), \end{align} which indeed vanishes as $R\rightarrow \infty$.

Answer 2

Recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty}x^{s-1}f(x)dx \implies F'(s) = \int_{0}^{\infty} x^{s-1}\ln(x)f(x) dx.$$

So, taking $f(x)=\frac{1}{x^4+1}$ and finding its Mellin transform

$$ F(s)=\frac{1}{4}\,{\frac {\pi }{\sin \left( \frac{\pi \,s}{4} \right) }} \implies F'(s)=\frac{1}{16}\,{\frac {{\pi }^{2}\cos \left( \pi \,s/4 \right) }{ \left( \cos \left( \pi \,s/4 \right) \right) ^{2}-1}}$$

Taking the limit as $s\to 3$ yields the desired result

$$ \lim_{s\to 3}F'(s)=\frac{\pi^2}{8\sqrt{2}}. $$

Answer 3

$$\underbrace{\dfrac{x^2 \log(x)}{1+x^4} dx \to \dfrac1{x^2} \dfrac{\log(1/x)}{1+1/x^4} \dfrac{-dx}{x^2}}_{x \to 1/x}$$ Hence, \begin{align} I = \int_0^{\infty} \dfrac{x^2 \log(x)}{1+x^4} dx & = \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx + \int_1^{\infty} \dfrac{x^2\log(x)}{1+x^4}dx\\ & = \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx - \int_0^1 \dfrac{\log(x)}{1+x^4}dx\\ & = \int_0^1 \dfrac{x^2-1}{1+x^4} \log(x) dx \end{align} \begin{align} \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx & = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{4k+2} \log(x)dx\\ & = \sum_{k=0}^{\infty}(-1)^k \int_{0}^{\infty} x^{4k+2} \log(x)dx\\ & = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(4k+3)^2} \end{align} \begin{align} \int_0^1 \dfrac{\log(x)}{1+x^4} dx & = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{4k} \log(x)dx\\ & = \sum_{k=0}^{\infty}(-1)^k \int_{0}^{\infty} x^{4k} \log(x)dx\\ & = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(4k+1)^2} \end{align} $$I = \sum_{k=0}^{\infty} (-1)^k \left(\dfrac1{(4k+1)^2} - \dfrac1{(4k+3)^2}\right) = \dfrac{\pi^2}{8\sqrt2}$$ which can be obtained from the identity that the PolyLogarithmic function satisfies $$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ You can see the posts here and here for more details.

Answer 4

Another way to handle these kinds of integrals is by setting $$ f(z) = \frac{z^2 \log^2 z}{1+z^4} $$ where $\log$ is the "natural branch" of the complex logarithm and integrate $f$ along a keyhole contour, see for example this. The idea with this approach is to take advantage of the branch cut of $\log$.

Answer 5

Use the substitution $x^4=t \Rightarrow dx=dt/(4t^{3/4})$ to obtain the following definite integral: $$\frac{1}{16}\int_0^{\infty} \frac{t^{-1/4}\ln t}{1+t}\,dt$$ Consider the result I proved here: Evaluate $\int_0^\infty\!\!\int_0^\infty\!\!\int_0^\infty\!\frac{(xyz)^{-1/7}(yz)^{-1/7}z^{-1/7}}{(x+1)(y+1)(z+1)}dx\,dy\,dz$ which states: $$\int_0^{\infty} \frac{t^{-a}}{1+t}\,dt=\frac{\pi}{\sin (a\pi)}$$ Differentiate the above with respect to $a$ to get: $$\int_0^{\infty} \frac{t^{-a}\ln t}{1+t}dt=\pi^2\cot(a\pi)\csc(a\pi)$$ Substitute $a=1/4$ and divide both the sides by 16 to get: $$\frac{1}{16}\int_0^{\infty} \frac{t^{-1/4}\ln t}{1+t}dt=\frac{\pi^2}{8\sqrt{2}}$$

Answer 6

We can use Integration under Differentiation to solve/generalize this problem instead of residues or special functions. It is much easier. Let $$ J_n(a)=\int_0^\infty \frac{x^a}{1+x^n}dx. $$ Clearly $$ J_n'(a)=\int_0^\infty \frac{x^2\ln x}{1+x^n}dx. $$ Note $J_4'(0)=I$. First we calculate \begin{eqnarray} J_n(a)&=&\int_0^1\frac{x^a+x^{n-a-2}}{1+x^n}dx\\ &=&\int_0^1\sum_{k=0}^\infty(-1)^n(x^a+x^{n-a-2})x^{kn}dx\\ &=&\sum_{k=0}^\infty(-1)^k\left(\frac{1}{nk+a+1}+\frac{1}{nk+n-a-1}\right)\\ &=&\sum_{k=0}^\infty(-1)^k\frac{1}{nk+a+1}\\ &=&\sum_{k=-\infty}^\infty(-1)^k\frac{1}{nk+a+1}\\ &=&\frac{\pi}{n\sin(\frac{a+1}{n}\pi)}. \end{eqnarray} So $$ J_n'(a)=-\frac{\pi^2\cos(\frac{(a+1)\pi}{n})}{n^2\sin^2(\frac{(a+1)\pi}{n})} $$ and hence $$ I=J_4'(2)=\frac{\pi^2}{8\sqrt2}. $$ Here we used the following result $$ \sum_{k=-\infty}^\infty(-1)^k\frac{1}{ak+b}=\frac{\pi}{a\sin(\frac{b}{a}\pi)}.$$

Answer 7

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x^{2}\ln\pars{x} \over 1 + x^{4}}\,\dd x:\ {\large ?}}$

The following method 'avoids' the $\ds{\large\it\underline{\mbox{four poles}}}$ of the original integral.

Instead, we just have to consider a $\ds{\large\it\underline{\mbox{one pole-integral}}}$:

Note that \begin{align}&\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{4}} \,\dd x} =\int_{0}^{\infty}{x^{1/2}\ln\pars{x^{1/4}} \over 1 + x}\,{1 \over 4}\,x^{-3/4}\,\dd x ={1 \over 16}\int_{0}^{\infty}{x^{-1/4}\ln\pars{x} \over 1 + x}\,\dd x \\[3mm]&={1 \over 16}\lim_{\mu\ \to\ -1/4}\partiald{}{\mu}\color{#00f}{% \int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}\tag{1} \end{align}

\begin{align} &\color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x} =2\pi\ic\expo{\pi\mu\ic} - \int_{\infty}^{0} {x^{\mu}\expo{2\pi\mu\ic} \over 1 + x}\,\dd x \\[3mm]&\imp\quad \color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x} =2\pi\ic\,{\expo{\pi\mu\ic} \over 1 - \expo{2\pi\mu\ic}} =-\pi\,{2\ic \over \expo{\pi\mu\ic} - \expo{-\pi\mu\ic}} =\color{#00f}{-\,{\pi \over \sin\pars{\pi\mu}}} \end{align}

Replacing in $\pars{1}$: \begin{align}&\color{#66f}{\large\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{4}} \,\dd x} ={1 \over 16}\bracks{\pi^{2}\cot\pars{\pi\mu}\csc\pars{\pi\mu}}_{\mu\ =\ -1/4} =\color{#66f}{\large {\root{2} \over 16}\,\pi^{2}} \approx -0.8724 \end{align}