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?
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$\begingroup$ Use beta function. You can get the logarithm by differentiation. $\endgroup$ – Zaid Alyafeai Dec 14 '14 at 11:19
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$\begingroup$ the result should be zero $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '14 at 11:24
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$\begingroup$ yes now is the result right $\frac{1}{8}\sqrt{2}\pi^2$ $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '14 at 11:52
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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}
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$\begingroup$ Nice! Thank you+1,It is said can use Double integral solve it $\endgroup$ – china math Dec 14 '14 at 12:46
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$\begingroup$ @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
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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.
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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
$$\int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} = \frac{\pi^2}{4 \sqrt{2}} $$
answered Dec 14 '14 at 22:42
