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I came across the following integral: $$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$ I know it could be solved with resuide method, and I want to know if there are some real methods can sove it? Meanwhile,I remember a similar integral: $$\large{\int_0^\infty \frac{1}{(x^2+1)(x^a+1)}\ dx=\frac{\pi}{4}}$$ And I want to know the following one: $${\color{red}{\large{\int_0^\infty \frac{\ln x}{(x^2+1)(x^a+1)}\ dx = \huge{?}}}}$$ Using the Mathematica I got the follow result. enter image description here

Could you suggest some ideas how to prove this? Any hints will be appreciated.

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Preliminary Result:

$\displaystyle \frac{2}{(x^2+1)(x^3+1)} \equiv \frac{x+1}{x^2+1}-\frac{x^2+x-1}{x^3+1}$

Proof: Obvious.

Consider the parametrised integral:

$\displaystyle f(\alpha) = \int \frac{x^\alpha \, \text{d}x}{(x^2+1)(x^3+1)} = \frac{1}{2} \int \frac{x^{\alpha+1}+x^\alpha}{x^2+1}-\frac{x^{\alpha+2}+x^{\alpha+1}-x^\alpha}{x^3+1} \, \text{d}x$

For the first part, substitute $t = x^2$, and $t = x^3$ for the second part

$\displaystyle f(\alpha) = \frac{1}{4} \int_0^\infty \frac{t^{\frac{\alpha}{2}}+t^\frac{\alpha-1}{2}}{1+t} \text{d}t +\frac{1}{6}\int_0^\infty \frac{t^{\frac{\alpha}{3}}+t^{\frac{\alpha-1}{3}}-t^{\frac{\alpha-2}{3}}}{1+t} \text{d}t$

Use the Integral Representations of the Beta Function to obtain:

$\displaystyle f(\alpha) = \frac{\text{B}(1+\frac{\alpha}{2},-\frac{\alpha}{2})+\text{B}(\frac{1+\alpha}{2},\frac{1-\alpha}{2})}{4} - \ \frac{\text{B}(1+\frac{\alpha}{3},-\frac{\alpha}{3})+\text{B}(\frac{2+\alpha}{3},\frac{1-\alpha}{3})-\text{B}(\frac{1+\alpha}{3},\frac{2-\alpha}{3})}{6}$

$\displaystyle \frac{f(\alpha)}{\pi} = \frac{\csc{\left(\pi+\frac{\pi \alpha}{2} \right)}+\csc{\left(\frac{\pi}{2}+\frac{\pi \alpha}{2}\right)}}{4}-\frac{\csc{\left( \pi + \frac{\pi \alpha}{3} \right)}+\csc{\left(\frac{2\pi}{3} +\frac{\pi \alpha}{3}\right)}-\csc{\left( \frac{\pi}{3} + \frac{\pi \alpha}{3}\right)}}{6}$

$\displaystyle \frac{12f(\alpha)}{\pi} = 3\sec{\frac{\pi \alpha}{2}} -3\csc{\frac{\pi \alpha}{2}}+2\csc{\frac{\pi \alpha}{3}}+2\csc{\left( \frac{2\pi}{3}+\frac{\pi \alpha}{3}\right)}-2\csc{\left(\frac{\pi}{3}+\frac{\pi \alpha}{3}\right)}$

Differentiate both sides with respect to $\alpha$ and evaluate at $0$ to obtain the integral you want.

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  • $\begingroup$ A very clear and straightforward solution. $\endgroup$ – poweierstrass Sep 28 '16 at 9:50
  • $\begingroup$ @Jack Lam Thank you Jack Lam for your excellent answer, I like your answer very much! $\endgroup$ – gcy-rolle Sep 28 '16 at 10:34
  • $\begingroup$ Nice answer (+1). i wanted to do the same but was too lazy to write a second answer $\endgroup$ – tired Sep 29 '16 at 10:33
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A real-analytic technique may be to exploit $$ I = \int_0^1 \frac{1-x^3}{(1+x^2)(1+x^3)}\log(x)\,dx $$ then expand $f(x)=\frac{1-x^3}{(1+x^2)(1+x^3)}$ as a Taylor series around $x=0$, $$ f(x) = \sum_{n\geq 0}\left(1-x^2-2 x^3+x^4+2 x^5+x^6-2 x^7-x^8 + x^{10} \right) x^{12n} $$ and exploit $$ \int_0^1 x^k\log(x)\,dx = -\frac{1}{(k+1)^2} $$ to convert $I$ into a combination of Dirichlet $L$-functions $L(s,\chi)$ where $s=2$ and $\chi$ is a Dirichlet character $\!\!\pmod{12}$, or a combination of trigamma functions evaluated at multiples of $\frac{1}{12}$: the reflection formula for $\psi'$ is helpful, since it gives $$ \sum_{n\geq 0}\left(\frac{1}{(12n+1)^2}+\frac{1}{(12n+11)^2}\right) = \frac{2+\sqrt{3}}{36}\,\pi^2,$$ $$ \sum_{n\geq 0}\left(\frac{1}{(12n+3)^2}+\frac{1}{(12n+9)^2}\right)=\frac{1}{72}\,\pi^2,$$ $$ \sum_{n\geq 0}\left(\frac{1}{(12n+4)^2}+\frac{1}{(12n+8)^2}\right)=\frac{1}{108}\,\pi^2,$$ $$ \sum_{n\geq 0}\left(\frac{1}{(12n+5)^2}+\frac{1}{(12n+7)^2}\right)=\frac{2-\sqrt{3}}{36}\,\pi^2,$$ $$ \sum_{n\geq 0}\frac{1}{(12n+6)^2}=\frac{1}{288}\pi^2.$$ To deduce $I=\color{red}{-\frac{37}{432}\pi^2}$ is now just a matter of simple algebra.

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    $\begingroup$ Wow, Thank you Jack D'Aurizio for your detailed answer. An excellent answer. $\endgroup$ – gcy-rolle Sep 27 '16 at 16:41
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    $\begingroup$ nice one but still cumbersome (as my method!). i wonder if something more elegant exists (+1) $\endgroup$ – tired Sep 27 '16 at 16:48
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Here is a variant of @tired's solution. In this solution, you are only required to know:

  • How to expand a rational function into partial fraction decomposition,
  • How to compute complex logarithm, and
  • The following claim:

Claim. For any $z \in \Bbb{C}\setminus(-\infty, 0]$ we have $$ I(z) := \int_{0}^{\infty} \left( \frac{\log x}{x+z} - \frac{\log x}{x+1} \right) \, dx = -\frac{1}{2}\log^2 z \tag{1}$$ where $\log$ is the principal value of the complex logarithm.


Computation. Before proving the claim, let us see how $\text{(1)}$ allows us to compute the integral in question. Using the partial fraction decomposition, we get

\begin{align*} \frac{1}{(1+x^2)(1+x^3)} &= \frac{1-i}{4} \cdot\frac{1}{x-i} + \frac{1+i}{4}\cdot\frac{1}{x+i} \\ &\qquad \qquad + \frac{1}{6}\cdot \frac{1}{x+1} - \frac{1}{3}\cdot \frac{1}{x - e^{\pi i/3}} - \frac{1}{3}\cdot \frac{1}{x - e^{-\pi i/3}}. \end{align*}

(Obviously this is a disguise of @tired's complex-analytic solution, since this decomposition often comes from residue computation. On the other hand, residue is no longer a necessity and thus one can work purely with algebra.)

Also notice that the sum of 'coefficients' are zero: $ \frac{1-i}{4} + \frac{1+i}{4} + \frac{1}{6} - \frac{1}{3} - \frac{1}{3} = 0$. Therefore it follows from the claim that

\begin{align*} \int_{0}^{\infty} \frac{\log x}{(1+x^2)(1+x^3)} \, dx &= \frac{1-i}{4} I(-i) + \frac{1+i}{4}I(i)\\ &\qquad \qquad + \frac{1}{6}I(1) - \frac{1}{3}I(-e^{\pi i/3}) - \frac{1}{3}I(-e^{-\pi i/3}) \\ &= -\frac{1}{2} \bigg[ \frac{1-i}{4} \left( -\frac{i\pi}{2} \right)^2 + \frac{1+i}{4}I\left( \frac{i\pi}{2} \right)^2 \\ &\qquad \qquad + 0 - \frac{1}{3}\left( -\frac{2i\pi}{3} \right)^2 - \frac{1}{3}\left( \frac{2i\pi}{3} \right)^2 \bigg] \\ &= -\frac{37}{432}\pi^2. \end{align*}


Proof of Claim. Now it remains to compute the integral $\text{(1)}$ This is easily done by differentiating $I(z)$:

$$ I'(z) = - \int_{0}^{\infty} \frac{\log x}{(x+z)^2} \, dx. $$

In order to compute this integral, we first replace the lower limit by $\epsilon > 0$ to obtain

$$ - \int_{\epsilon}^{\infty} \frac{\log x}{(x+z)^2} \, dx = \frac{\epsilon \log \epsilon}{z(z+\epsilon)} - \frac{1}{z}\log (z+\epsilon). $$

Taking limit as $\epsilon \to 0^+$, we get

$$ I'(z) = -\frac{1}{z}\log z, \qquad I(1) = 0. $$

This is enough to prove the claim. ////

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    $\begingroup$ Thank you very much for your excellent answer. $\endgroup$ – gcy-rolle Sep 29 '16 at 10:09
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Let's define the branch of $\log$ to lie on the posive real axis. We consider the complex valued function

$$ f(z)=\frac{\log(z)^2}{(z^2+1)(z^3+1)} $$

we integrate this function around a keyhole contour in the complex plane with slit at the positive real axis. The contributions from the big circle will vanish out because $f(|z|)\sim \log(|z|)/|z|^5$ as $|z|\rightarrow \infty$ so we are left with

$$ \oint f(z)=\int_0^{\infty}\frac{\log(x)^2}{(x^2+1)(x^3+1)}-\int_0^{\infty}\frac{(\log(x)+2 i \pi)^2}{(x^2+1)(x^3+1)}=2\pi i\sum_i\text{Res}(f(z),z=z_i) $$

where are the roots of the denominator of $f(z)$: $z_0=-1,z_{2/3}=\pm i,z_{4,5}=e^{\pm i2 \pi/3}$. We can rewrite this as $$ \int_0^{\infty}\frac{\log(x)}{(x^2+1)(x^3+1)}=\sum_i\text{Res}(f(z),z=z_i)-2\pi i\int_0^{\infty}\frac{1}{(x^2+1)(x^3+1)} $$

The remaining integral can be evaluated either by cumbersome partial fraction decomposition or by considering

$$ g(z)=\frac{\log(z)}{(z^2+1)(z^3+1)} $$

integrated around the same contour then above.

I leave the details to you but from now on (also the calculation of residues if one keep in mind $\log(i)=\frac{i\pi}{2},\log(i)=\frac{3i\pi}{2}$ for this choice of branchcut) so i leave the details to you...

As an alternative the remaining integral might be ignored because it will yield a purely imaginary contribution and we can write (the lhs is purely real)

$$ \int_0^{\infty}\frac{\log(x)}{(x^2+1)(x^3+1)}=\Re\left[\sum_i\text{Res}(f(z),z=z_i)\right] $$


This method will work for

$$ \int_0^{\infty}\frac{\log(z)^2}{(z^2+1)(z^n+1)}dz $$

as long $n$ is a natural number

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    $\begingroup$ (+1) It is a perfectly sound method, but not a real method :D $\endgroup$ – Jack D'Aurizio Sep 27 '16 at 16:43
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    $\begingroup$ @Jack D'Aurizio this question is tagged as "residue calculus " soi have no remorse :D. Thx anyway! $\endgroup$ – tired Sep 27 '16 at 16:44
  • $\begingroup$ @tired A very nice answer. For the integral of $\ln x$,the residue method always to be effective. And I'm sorry for tagging as "residue calculus". $\endgroup$ – gcy-rolle Sep 27 '16 at 16:49
  • $\begingroup$ @Downvoter...have you cared to read the comments under my answer? $\endgroup$ – tired Sep 27 '16 at 16:59
  • $\begingroup$ @ tired Ok,I will do it now. $\endgroup$ – gcy-rolle Sep 27 '16 at 17:01

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