4
$\begingroup$

How to evaluate the following integral $$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$ without using residue or complex analysis methods?

$\endgroup$
3
  • $\begingroup$ Use beta function. You can get the logarithm by differentiation. $\endgroup$ – Zaid Alyafeai Dec 14 '14 at 11:19
  • $\begingroup$ the result should be zero $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '14 at 11:24
  • $\begingroup$ yes now is the result right $\frac{1}{8}\sqrt{2}\pi^2$ $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '14 at 11:52
12
$\begingroup$

We have a well-known formula below $$J(a,b)=\int_0^\infty\frac{x^{\large a-1}}{1+x^b}\mathrm dx=\frac{\pi}{b}\csc\left(\frac{a\pi}{b}\right)\tag{1}$$ Differentiating $(1)$ with respect to $a$ once, we have $$J'(a,b)=\int_0^\infty\dfrac{x^{\large a-1}\ln x}{1+x^b}\mathrm dx=-\frac{\pi^2}{b^2}\csc\left(\frac{a\pi}{b}\right)\cot\left(\frac{a\pi}{b}\right)\tag{2}$$ then, by using $(2)$, we can obtain the result of our integral as follows \begin{align} I&=\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}\mathrm dx\\[10pt] &=\int_{0}^{\infty}\frac{x^2\;\ln{x}}{1+x^4}\mathrm dx-\int_{0}^{\infty}\frac{\ln{x}}{1+x^4}\mathrm dx\\[10pt] &=J'(3,4)-J'(1,4)\\[10pt] &=\frac{\pi^2}{8\sqrt{2}}+\frac{\pi^2}{8\sqrt{2}}\\[10pt] &=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi^2}{4\sqrt{2}}}} \end{align}

$\endgroup$
3
  • $\begingroup$ Nice! Thank you+1,It is said can use Double integral solve it $\endgroup$ – china math Dec 14 '14 at 12:46
  • $\begingroup$ @chinamath You're welcome. Let me think that way first. If I can find it, I'll post the new one $\endgroup$ – Venus Dec 14 '14 at 12:49
  • $\begingroup$ I think we must take $\ln{x}=\int (?)$ $\endgroup$ – china math Dec 14 '14 at 12:52
4
$\begingroup$

Noting that $$ \int_0^1x^n\ln x\,dx=-\frac{1}{(n+1)^2} $$ we have \begin{eqnarray} \int_0^\infty\frac{(x^2-1)\ln x}{1+x^4}dx&=&2\int_0^1\frac{(x^2-1)\ln x}{1+x^4}\,dx\\ &=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^2-1)x^{4n}\ln x\,dx\\ &=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n+2}-x^{4n})\ln x\,dx\\ &=&2\sum_{n=0}^\infty(-1)^n\left(\frac1{(4n+1)^2}-\frac1{(4n+3)^2}\right)\\ &=&2\sum_{n=-\infty}^\infty(-1)^n\frac1{(4n+1)^2}\\ &=&\frac{1}{32}\left(\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{8})^2}-\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{8})^2}\right)\\ &=&\lim_{b\to0}\frac{1}{32}\left(\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}\bigg|_{a=-\frac{1}{8}}-\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}\bigg|_{a=-\frac{3}{8}}\right)\\ &=&\frac{\pi^2}{4\sqrt2}. \end{eqnarray} Here we use this.

$\endgroup$
4
  • $\begingroup$ Simply Nice ! :-) $\endgroup$ – r9m Dec 16 '14 at 9:45
  • $\begingroup$ how did you get from the integral $(0,\infty)$ to $(0,1)$? $\endgroup$ – farruhota Dec 23 '20 at 16:37
  • 1
    $\begingroup$ Split $(0,\infty)$ to $(0,1)$ and $(1,\infty)$. For $(1,\infty)$, change variable$x\to\frac1x$ and then you will get it. $\endgroup$ – xpaul Dec 23 '20 at 17:05
  • $\begingroup$ Excellent. Thank you. +1 $\endgroup$ – farruhota Dec 24 '20 at 16:45
2
$\begingroup$

Consider the contour integral

$$\oint_C dz \frac{(z^2-1) \log^2{z}}{1+z^4} $$

where $C$ is a keyhole contour about the positive real axis having an outer radius $R$ and an inner radius $\epsilon$. As $R \to \infty$ and $\epsilon \to 0$, the integral may be shown to be equal to

$$-i 4 \pi \int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} + 4 \pi^2 \int_0^{\infty} dx \frac{x^2-1}{1+x^4} $$

The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, which are at $e^{i (2 k+1) \pi/4}$ for $k=0,1,2,3$, or

$$i \frac{\pi}{2} \left [\frac{(i-1) (-\pi^2/16)}{e^{i 3 \pi/4}} - \frac{(i+1) (-9\pi^2/16)}{e^{i \pi/4}} + \frac{(i-1) (-25 \pi^2/16)}{e^{-i \pi/4}} - \frac{(i+1) (-49\pi^2/16)}{e^{-i 3 \pi/4}} \right ]$$

which simplifies to $-i (\pi^3/32) 16 \sqrt{2} = -i \pi^3/\sqrt{2}$. Equating real and imaginary parts, we find that

$$\int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} = \frac{\pi^2}{4 \sqrt{2}} $$

$\endgroup$

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.