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Solve $$I=\int_{0}^{\infty}\frac{\ln^2x}{1+x^4}dx$$

Using function Euler. Help me please..

Narasimham
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K.D.H.D.
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    What is "function Euler"? Are you thinking of a particular programming language and library? Also, is that an iterated logarithm, or a power in the integrand? – parsiad May 03 '16 at 20:48
  • What does Euler's Toitient have to do with evaluation of this integral? Or perhaps you meant something different from that. – Mark Viola May 03 '16 at 21:35
  • By breaking the original integration range into $(0,1]\cup[1,+\infty)$ and performing the trivial substitution, the integral takes the form $$\int_{0}^{1}\frac{1+x^2}{1+x^4}\log^2(x),dx.$$ By expanding $\frac{1+x^2}{1+x^4}$ as a Taylor series around $x=0$ and exploiting $\int_{0}^{1}x^n \log(x)^2,dx = \frac{2}{(n+1)^3}$, the problem can be solved through a well-known identity: http://math.stackexchange.com/questions/850442/an-interesting-identity-involving-powers-of-pi-and-alternating-zeta-series – Jack D'Aurizio May 03 '16 at 23:15

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You can use contour integration here: \begin{align*} \int_0^{\infty} \frac{\ln(x)^2 + (\ln(x) + \pi i)^2}{1+x^4} \, \mathrm{d}x &= \int_{-\infty}^{\infty} \frac{\log(z)^2}{1+z^4} \, \mathrm{d}z \\ &= 2\pi i \mathrm{Res} \Big( \frac{\log(z)^2}{1+z^4}; z=\sqrt{i} \Big) + 2\pi i \mathrm{Res} \Big( \frac{\log(z)^2}{1+z^4}; z= \sqrt{i}^3 \Big) \\ &=2 \pi i \Big( -\frac{\log(\sqrt{i})^2 \sqrt{i}}{4} - \frac{\log(\sqrt{i}^3)^2 \sqrt{i}^3}{4}\Big) \\ &= 2\pi i \Big(\frac{(\pi i / 4)^2 (-1-i)}{4 \sqrt{2}} + \frac{(3\pi i / 4)^2 (1 - i)}{4 \sqrt{2}}\Big) \\ &= -\frac{\pi^3}{4\sqrt{2}}i - \frac{5 \pi^3}{16 \sqrt{2}}, \end{align*} so $$2 \int_0^{\infty}\frac{\ln(x)^2}{1+x^4} \, \mathrm{d}x + 2\pi i \int_0^{\infty} \frac{\ln(x)}{1+x^4} - \pi^2 \int_0^{\infty} \frac{1}{1+x^4} \, \mathrm{d}x = -\frac{\pi^3}{4\sqrt{2}}i - \frac{5 \pi^3}{16 \sqrt{2}}.$$ Taking real parts, $$\int_0^{\infty} \frac{\ln(x)^2}{1+x^4} = \frac{\pi^2}{2} \int_0^{\infty} \frac{1}{1+x^4} \, \mathrm{d}x - \frac{5\pi^3}{32 \sqrt{2}}$$ Another contour integral will calculate $\int_0^{\infty} \frac{1}{1+x^4} \, \mathrm{d}x = \frac{\pi}{2 \sqrt{2}}$, so $$\int_0^{\infty} \frac{\ln(x)^2}{1+x^4} \, \mathrm{d}x = \frac{\pi^3}{4 \sqrt{2}} - \frac{5\pi^3}{32 \sqrt{2}} = \frac{3 \pi^3}{32 \sqrt{2}}.$$

user336784
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