This integral below $$ I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2} $$ is what I am trying to prove. Thanks.

We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$. Next I wrote $$ I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\ \int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx, $$ however only the middle integral is convergent. I am not sure how to go about solving this problem. Thank you

  • 1
    $\begingroup$ The easiest way to solve it is using complex analysis . $\endgroup$ – Zaid Alyafeai May 12 '14 at 19:17
  • 1
    $\begingroup$ @ZaidAlyafeai: Mellin transform technique is easier! $\endgroup$ – Mhenni Benghorbal May 12 '14 at 19:24
  • $\begingroup$ @MhenniBenghorbal, I should never use the term easy because it is slippery. $\endgroup$ – Zaid Alyafeai May 13 '14 at 20:20
  • $\begingroup$ @ZaidAlyafeai: you are the one who used it in your comment! So that's what you think and I wrote down what I think :). $\endgroup$ – Mhenni Benghorbal May 13 '14 at 20:28

Following Mhenni's suggestion, I will calculate $$I(\mu) = \int_0^{\infty} x^{\mu}\frac{1+x^2}{1+x^4} \,dx $$ and then take $I''(0)$. By the ubiquitous formula $$\int_0^{\infty}\frac{x^a}{1+x^b} \,dx =\frac{\pi}{b \sin(\pi(a+1)/b)}$$ we obtain $$I(\mu)=\frac{\pi}{4} \left[ \frac{1}{\sin(\pi(\mu+1)/4)} + \frac{1}{\sin(\pi(\mu+3)/4)} \right]$$ This gives (after some simple algebra)

$$I''(0) = \int_0^{\infty} \frac{1+x^2}{1+x^4} \log^2 x \,dx = \frac{\pi}{4}\left[\frac{3 \pi ^2}{8 \sqrt{2}}+\frac{3 \pi ^2}{8 \sqrt{2}} \right] = \frac{3 \pi ^3}{16 \sqrt{2}}$$ as was to be shown.

  • 2
    $\begingroup$ @MhenniBenghorbal Basically yes, only I bothered to write it down for him :P $\endgroup$ – user111187 May 13 '14 at 4:20
  • 1
    $\begingroup$ @user111187: We expect the OP to do some work! $\endgroup$ – Mhenni Benghorbal May 13 '14 at 5:54
  • 3
    $\begingroup$ @MhenniBenghorbal I already have a solution to this problem. It is nice to see others work out details to different methods, considering you never provide a complete solution to any problem. Merely, you post a bunch of related problems. $\endgroup$ – Jeff Faraci May 13 '14 at 19:48
  • 1
    $\begingroup$ @MhenniBenghorbal I do plenty of work, much more work regarding these integrals than you do, as is clearly shown. $\endgroup$ – Jeff Faraci May 13 '14 at 19:59
  • 2
    $\begingroup$ @Integrals:Good for you! Keep the hard work. I am happy to assist or share some ideas. $\endgroup$ – Mhenni Benghorbal May 13 '14 at 20:30

A related problem. Recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x)dx \implies F''(s)=\int_{0}^{\infty} \ln(x)^2x^{s-1}f(x)dx .$$

Now the whole problem boils down to finding the Mellin transform of $\frac{1+x^2}{1+x^4}$, differentiating twice, and then taking the limit as $s \to 1$.

Can you finish it?


Here is another way to calculate. It is much simpler. Clearly \begin{eqnarray*} I&=&2\int_0^1 \log^2 x\frac{1+x^2}{1+x^4}dx\\ &=&2\int_0^1\sum_{n=0}^\infty(1+x^2)(-1)^nx^{4n}\log^2xdx\\ &=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\ &=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\ &=&4\sum_{n=0}^\infty(-1)^n(\frac{1}{(4n+1)^3}+\frac{1}{(4n+3)^3})\\ &=&4\sum_{n=-\infty}^\infty(-1)^n\frac{1}{(4n+1)^3}\\ &=&\frac{3\pi^3}{16\sqrt2} \end{eqnarray*} Here we use the following theorem $$ \sum_{n=-\infty}^\infty (-1)^nf(n)=-\pi \sum_{k=1}^m\text{Re}(\frac{f(z)}{\sin\pi z},a_k) $$ where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$. For $f(z)=\frac{1}{(4z+1)^3}$, $z=-\frac{1}{4}$ is the only pole and $$ \text{Re}(\frac{f(z)}{\sin\pi z},-\frac{1}{4})=-\frac{3\pi^2}{64\sqrt2}. $$ Thus we have the result.


$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I\equiv\int_{0}^{\infty}\ln^{2}\pars{x}\,{1 + x^{2} \over 1 + x^{4}}\,\dd x ={3 \pi^{3} \over 16 \root{2}}}$

\begin{align} I&=-\Im\bracks{\pars{1 - \ic} \int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + \ic}\,\dd x} \\[3mm]&=-\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty} x^{\mu}\int_{0}^{\infty}\expo{-\pars{x^{2} + \ic}\xi}\,\dd\xi\,\dd x} \\[3mm]&=-\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty}\expo{-\ic\xi}\ \overbrace{\int_{0}^{\infty}x^{\mu}\expo{-\xi x^{2}}\,\dd x}^{\ds{t \equiv \xi x^{2}\ \imp\ x = \xi^{-1/2}t^{1/2}}}\ \dd\xi} \\[3mm]&=-\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty}\expo{-\ic\xi}\ \int_{0}^{\infty}\xi^{-\mu/2}\ t^{\mu/2}\expo{-t}\xi^{-1/2}\,\half\,t^{-1/2}\,\dd t\, \dd\xi} \\[3mm]&=-\,\half\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty} \xi^{-\pars{\mu + 1}/2}\expo{-\ic\xi} \int_{0}^{\infty}t^{\pars{\mu - 1}/2}\expo{-t}\,\dd t\,\dd\xi} \\[3mm]&=-\,\half\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\Gamma\pars{{\mu \over 2} + \half} \color{#c00000}{\int_{0}^{\infty}\xi^{-\pars{\mu + 1}/2}\expo{-\ic\xi}\,\dd\xi}} \tag{1} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

Also, \begin{align} &\overbrace{% \color{#c00000}{\int_{0}^{\infty}\xi^{-\pars{\mu + 1}/2}\expo{-\ic\xi}\,\dd\xi}} ^{\ds{t \equiv \ic\xi\quad\imp\quad\xi = -\ic t = \expo{-\ic\pi/2}t}} =\int_{0}^{\infty\ic}\pars{\expo{-\ic\pi/2}t}^{-\pars{\mu + 1}/2} \expo{-t}\,\pars{-\ic\,\dd t} \\[3mm]&=-\ic\expo{\ic\pi\pars{\mu + 1}/4}\int_{0}^{\infty}t^{-\pars{\mu + 1}/2} \expo{-t}\,\dd t=-\ic\expo{\ic\pi\pars{\mu + 1}/4}\Gamma\pars{\half - {\mu \over 2}} \end{align}

Expression $\pars{1}$ is reduce to: \begin{align} I&=-\,\half\,\Im\braces{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\Gamma\pars{{\mu \over 2} + \half} \bracks{-\ic\expo{\ic\pi\pars{\mu + 1}/4}\Gamma\pars{\half - {\mu \over 2}}}} \\[3mm]&=\half\,\Im\bracks{\pars{1 + \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\expo{\ic\pi\pars{\mu + 1}/4}\, {\pi \over \sin\pars{\pi\bracks{\mu/2 + 1/2}}}} \end{align} where we used Euler Reflection Formula ${\bf\mbox{6.1.17}}$.

\begin{align} I&={\root{2} \over 2}\,\pi\, \lim_{\mu \to 0}\partiald[2]{}{\mu}\bracks{\cos\pars{\pi\mu \over 4} \sec\pars{\pi\mu \over 2}} ={\pi \over \root{2}}\pars{-\,{\pi^{2} \over 16} + {\pi^{2} \over 4}} \end{align}

$$\color{#00f}{\large% I\equiv\int_{0}^{\infty}\ln^{2}\pars{x}\,{1 + x^{2} \over 1 + x^{4}}\,\dd x ={3 \pi^{3} \over 16 \root{2}}} $$

  • $\begingroup$ @Integrals I was trying to find a different answer but it became too long. Thanks. $\endgroup$ – Felix Marin May 13 '14 at 20:26

The integrand is invariant under inversion:



$$\begin{align} \int_0^\infty \log^2x{1+x^2\over1+x^4}dx&=\int_0^1 \log^2x{1+x^2\over1+x^4}dx+\int_1^\infty \log^2x{1+x^2\over1+x^4}dx\\ &=\int_0^1 \log^2x{1+x^2\over1+x^4}dx-\int_1^0 \log^2x{1+x^2\over1+x^4}dx\\ &=2\int_0^1 \log^2x{1+x^2\over1+x^4}dx \end{align}$$

Maye now you can expand things as a power series.

  • $\begingroup$ One can simmilarly show that $$ I=2\int_0^1 (\log x)^2{1+x^2\over1+x^4}\mathrm{d}x =4\int_0^1(\log x)^2{x^2\over1+x^4}\mathrm{d}x =4\int_0^1(\log x)^2{\mathrm{d}x \over1+x^4} $$ By splitting the integrals again and use inversion $1/x$. We have $\sum_{n\geq 0}(-x^4)^n = 1/(1+x^4)$ so $$ I = \sum_{n\geq1} \int_0^1 (\log x)^2(-x^4)^n = \sum_{n\geq1}\frac{2(-1)^n}{64n^3 + 48n^2+12n+1} $$ But this seems harder to evaluate than the original integral. I think it is wiser to do the $\frac{(\log x)^2}{1+x^4}$ throgh complex analysis, or other clever means. $\endgroup$ – N3buchadnezzar May 12 '14 at 19:57
  • $\begingroup$ @N3buchadnezzar, you can't use inversion again, because doing so takes the limits of integration from $(0,1)$ back to $(1,\infty)$. $\endgroup$ – Barry Cipra May 12 '14 at 20:10
  • $\begingroup$ @N3buchadnezzar, but I agree that the infinite series you get, which I find to be $$4(1+{1\over3^3}-{1\over5^3}-{1\over7^3}+{1\over9^3}+{1\over11^3}-\cdots)$$ still has to be shown equal to $3\pi^3/16\sqrt2$, which ain't obvious. $\endgroup$ – Barry Cipra May 12 '14 at 20:14
  • $\begingroup$ @N3buchadnezzar, see the answer below. $\endgroup$ – xpaul Jul 29 '14 at 21:18

A little long answer, but let us consider the triple integral:

$$I=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2)(1+y^2)(1+x^2y^2z^4)} \ dz \ dy \ dx.$$

Integrating in this order, we use a u substitution $z=\frac{u}{\sqrt{xy}}$ and $\ dz =\frac{du}{\sqrt{xy}},$ and this transforms the integral into

$$I=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{\sqrt{xy}\left(1+\frac{u^2}{xy} \right)}{(1+x^2)(1+y^2)(1+u^4)} \ du \ dy \ dx.$$ We can apply the well-known integral formula $$\int_{0}^{\infty} \frac{v^m}{1+v^n} \ dv =\frac{\pi}{n} \csc \left(\frac{\pi(m+1)}{n} \right)$$ to get that $$I=\frac{\pi^3}{2\sqrt{2}}.$$ Now let us reverse the order of integration as such.

$$I=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2)(1+y^2)(1+x^2y^2z^4)} \ dy \ dz \ dx.$$ Using partial fractions or Mathematica, we can get that

$$I= \int_{0}^{\infty}\int_{0}^{\infty} \frac{x(1+z^2)\ln(x)}{(1+x^2)(x^2z^4-1)} \ dz \ dx+\int_{0}^{\infty}\int_{0}^{\infty} \frac{2x(1+z^2)\ln(z)}{(1+x^2)(x^2z^4-1)} \ dz \ dx = J_1 + J_2.$$ Reverse the order of integration in $J_2$ and use partial fractions to see

$$J_2=4\int_{0}^{\infty} \frac{(z^2+1)(\ln(z))^2}{z^4+1} \ dz.$$ Let's focus on $J_1.$

Integrating with respect to $z$ first with partial fractions gives

$$J_1= \int_{0}^{\infty} \frac{\pi}{4\sqrt{x}} \frac{(1-x)\ln(x)}{1+x^2} \ dx,$$ and we expand $$J_1=-\int_{0}^{\infty}\int_{0}^{\infty} \frac{\pi(1-x)}{4\sqrt{x}} \frac{t}{(1+t^2)(1-t^2x^2)} \ dt \ dx,$$ which can be proven by partial fractions.

Reverse the order of integration and use partial fractions again to get that

$$J_1= -\frac{\pi^2}{8} \int_{0}^{\infty} \frac{t+1}{\sqrt{t}(t^2+1)} \ dt,$$ which by the well-known formula I mentioned, can be shown that $$J_1=\frac{-\pi^3}{4\sqrt{2}}.$$

Putting everything together now, we have that

$$\frac{\pi^3}{2\sqrt{2}}=\frac{-\pi^3}{4\sqrt{2}}+4\int_{0}^{\infty} \frac{(z^2+1)(\ln(z))^2}{z^4+1} \ dz,$$ which by some algebraic manipulation, gives us

$$\int_{0}^{\infty} \frac{(z^2+1)(\ln(z))^2}{z^4+1} \ dz= \frac{3 \pi^3}{16\sqrt{2}}.$$

  • 1
    $\begingroup$ Yikes,It's painful looking at some of these calculations.It makes you sympathetic for mathematics and science in the days before modern computers. $\endgroup$ – Mathemagician1234 Aug 1 '16 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.