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How to evaluate these two integrals below $$\int_{0}^{\infty }\frac{\ln\left ( 1+x^{4} \right )}{1+x^{2}}\mathrm{d}x$$ $$\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}\mathrm{d}x$$ For the first one, I tried to use $$\mathcal{I}'(s)=\int_{0}^{\infty }\frac{x^4}{(1+sx^4)(1+x^{2})}\mathrm{d}x$$ but it seems hard to solve.

Quanto
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Renascence_5.
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  • It is easy to solve by contour integration – Eugene Zhang Mar 08 '16 at 05:52
  • @hermes I'm not familiar with contour integration – Renascence_5. Mar 08 '16 at 05:59
  • Without contour integration, it could be hard. You can try taylor series – Eugene Zhang Mar 08 '16 at 06:02
  • @EvilNebula. could u please start to give some context and solution attempts in the (often interesting) questions u are asking? Otherwise they will shot down frequently – tired Mar 08 '16 at 08:33
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    I don't understand why this question was closed: if the asker doesn't have any real idea how to solve, and he even wrote an attempt, as scarce as it was, then he has no right to ask? I think this isn't correct in this case. – DonAntonio Mar 08 '16 at 09:37

3 Answers3

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For the first integral, we introduce a new parameter $a$. Let $I(a)$ be the integral

$$I(a)=\int_0^\infty \frac{\log(a+x^4)}{1+x^2}\,dx \tag 1$$

We note that it can be shown that $I(0)=0$. Our integral of interest is $I(1)$.

Proceeding, we differentiate $(1)$ to obtain

$$\begin{align} I'(a)&=\int_0^\infty\frac{1}{(1+x^2)(a+x^4)}\,dx\\\\ &=\frac{1}{2(a+1)}\int_{-\infty}^\infty\left(\frac{1}{x^2+1}+\frac{1-x^2}{x^4+a}\right)\,dx\\\\ &= \frac{2\pi i}{2(a+1)}\left(\frac{1}{2i}+\frac{1}{i\,2^{3/2}a^{3/4}}-\frac{1}{i\,2^{3/2}a^{1/4}}\right)\\\\ &=\frac{\pi}{2(1+a)}+\frac{\pi}{2^{3/2}(1+a)a^{3/4}}-\frac{\pi}{2^{3/2}(1+a)a^{1/4}}\tag 2 \end{align}$$

Integrating the right-hand side of $(2)$ reveals

$$\begin{align} I(1)&=\frac{\pi}{2}\log(2)+\frac{\pi}{2^{3/2}}\int_0^1\left(\frac{1}{(1+x)x^{3/4}}-\frac{1}{(1+x)x^{1/4}}\right)\,dx\\\\ &=\frac{\pi}{2}\log(2)+\pi\sqrt 2\int_0^1\frac{1-x^2}{1+x^4}\,dx\\\\ &=\frac{\pi}{2}\log(2)+\pi\sqrt 2\left.\left(\frac{\log(x^2+\sqrt2 x+1)-\log(-x^2+\sqrt 2x-1)}{2^{3/2}}\right)\right|_{0}^{1}\\\\ &=\frac{\pi}{2}\log(2)+\frac{\pi}{2}\log\left(\frac{2+\sqrt 2}{2-\sqrt 2}\right)\\\\ &=\pi\log(2+\sqrt 2) \end{align}$$

Mark Viola
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Express the 2nd integral in terms of the 1st \begin{align} &\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x =\int_{0}^{1}\arctan x \ d\bigg(\frac12\ln \frac {x^2} {{1+x^4}} \bigg)\\ =&-\frac\pi8\ln2 +\frac12\int_0^1 \frac{\ln\frac{1+x^4}{x^2}}{1+x^2}\overset{x\to 1/x}{dx} =-\frac\pi8\ln2 +\frac14 \int_0^\infty \frac{\ln(1+x^4)}{1+x^2}dx \end{align}

Quanto
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Using the identity $\ln \left(x^2+y^2\right)=2 \operatorname{Re}(\ln (x+y i))$ to reduce the power $4$ of $x$ in the numerator makes the life easier. $$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x = & \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x \\ = & \operatorname{Re}\left(\int_{-\infty}^{\infty} \frac{\ln \left(x^2+i\right)}{1+x^2}\right)\\=&\operatorname{Re} \int_{-\infty}^{\infty} \frac{\ln \left(x^2+\left(\frac{1+i}{\sqrt{2}}\right)^2\right)}{1+x^2} \end{aligned} $$ Using the particular case $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+a^2\right)}{1+x^2} d x=2 \pi \ln (1+a)$ in the post, we have $$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x & =\operatorname{Re}\left[2 \pi \ln \left(1+\frac{1+i}{\sqrt{2}}\right)\right] \\ & =2 \pi \operatorname{Re}\left[\ln \left(1+\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]\\ & =\pi \ln \left(1+\frac{1}{2}+\sqrt{2}+\frac{1}{2}\right) \\ & =\pi \ln (2+\sqrt{2}) \end{aligned} $$

Lai
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