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Problem - S. Stewart (Saudi Arabia) Evaluate [I=\int_0^\infty\frac{\ln\left(\cos^2(x)\right)\sin^3(x)}{x^3\left(1+2\cos^2(x)\right)}\,\mathrm dx.] Solution We show that [I=-\frac{\pi}{4}\ln(2)...
Problem - B. Bradie (USA) Evaluate [I=\int_{-1}^1\frac{\text{arccos}(x)}{x^2+x+1}\,\mathrm dx.] Solution We show that [I=\frac{4\pi^2}{3\sqrt{3}}-\frac{2\pi}{\sqrt{3}}\text{arctan}\left(\sqrt{...
Problem - I. Mezo (China) Prove [\int_0^\infty\frac{\cos(x)-1}{x(e^x-1)}\,\mathrm dx=\frac{1}{2}\ln\left(\frac{\pi}{\sinh(\pi)}\right).] Solution First we acquire two preliminary identities. In...
Problem - F. Holland (Ireland) Prove [\int_0^{\infty}\left(\frac{\text{tanh}(x)}{x}\right)^2\,\mathrm dx=\frac{14\zeta(3)}{\pi^2}.] Solution Consider the contour integral [I_k=\oint_{C_k}\left...
Problem - S. Stewart (Saudi Arabia) Prove [\int_0^{\pi/2}\frac{\sin(4x)}{\ln(\tan(x))}\,\mathrm dx=-\frac{14\zeta(3)}{\pi^2}.] Solution We define [I(s)=\int_0^{\pi/2}\frac{\sin(4x)\tan^s(x)}{\...
The integral [\int_0^\infty\frac{\ln(x)}{(1+x)(1+x^2)}\,\mathrm dx] was brought to me by a friend, who came about it in a context that suggested for it to be done by contour integration. Surpsisi...
Problem Evaluate [\sum_{n=1}^\infty\text{arctan}\left(\frac{1}{8n^2}\right).] Solution We can compute this sum by considering it as the argument of an infinite product. Here the argument of the...
Problem - H. Ohtsuka (Japan) For $s>1$, prove the following inequalities: [\sum_p\frac{1}{p^s-\frac{1}{2}}<\log(\zeta(s)),\,\sum_p\frac{1}{p^s}<\log\left(\frac{\zeta(s)}{\sqrt{\zeta(2s)}...
Problem - Li Zhou (USA) For a nonnegative integer m, let [A_m=\sum_{k=0}^\infty\left(\frac{1}{(6k+1)^{2m+1}}-\frac{1}{(6k+5)^{2m+1}}\right).] Prove $A_0=\pi\sqrt{3}/6$ and, for $m\geq1$, [2A_m+...
Problem - P. Bracken (USA) Prove [\int_0^1\frac{\ln(1+x)\ln(1-x)}{x}\,\mathrm dx=-\frac{5\zeta(3)}{8}.] Solution Call the posed integral $I$ and define [J=\int_0^1\frac{1}{x}\ln^2\left(\frac{1...
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