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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

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    $\begingroup$ The easiest way to solve it is using complex analysis . $\endgroup$ – Zaid Alyafeai May 12 '14 at 19:17
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    $\begingroup$ @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
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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.

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    $\begingroup$ @MhenniBenghorbal Basically yes, only I bothered to write it down for him :P $\endgroup$ – user111187 May 13 '14 at 4:20
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    $\begingroup$ @user111187: We expect the OP to do some work! $\endgroup$ – Mhenni Benghorbal May 13 '14 at 5:54
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ @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
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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?

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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.

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

\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}

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

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

$$\log^2(1/x){1+(1/x)^2\over1+(1/x)^4}d(1/x)=-\log^2x{1+x^2\over1+x^4}dx$$

Thus

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

Maye now you can expand things as a power series.

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  • $\begingroup$ 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
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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

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

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    $\begingroup$ 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

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