# Tag Archives: logarithm

$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx = \frac{\pi^3}{12}+\pi \log^2(2)$$ Lemma $$\int^\infty_0 \frac{\log^3(1 + x^2)}{x^2}\,dx = \pi^3+ 3 \pi \log^2(4)$$ Start by the following $$\int^{\infty}_0 x^{-p}(1+x)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)}$$ Let $x\to x^2$ $$\int^{\infty}_0 x^{-2p+1}(1+x^2)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{2\Gamma(1-s)}$$ Let $p = 3/2$ $$\int^{\infty}_0 \frac{1}{x^2(1+x^2)^{1-s}} … Continue reading Posted in Beta function, Contour Integration | | 4 Comments ## Contour method for shifted logarithm branch Prove $a,b,c,d >0$$$\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+d^2x^2}\,dx = \frac{\pi}{cd} \log \frac{ad+bc}{d}$$Consider the function$$f(z) = \frac{\log(a-ibz)}{c^2+d^2z^2}$$We need the logarithm with the branch cut $y<-\frac{a}{b} , x =0$ . Note that this corresponds to$$\log(a+ibz) = \log\sqrt{(a+y)^2+b^2x^2}+i\theta … Continue reading

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