Sovle integral: ∫0∞log⁡(1+x2)coshπx2sinh2πx2dx=2−4π

Question

Sovle integral:  ∫0∞log⁡(1+x2)coshπx2sinh2πx2dx=2−4π

chumants6g · Accepted Answer

Integrating by parts shows that the integral is equivalent to showing that  ∫0∞x1+x21sinhπx2dx=π2−1  Let f(z)=z1+z21sinhπz2 and integrate around a rectangle with vertices at ±N and ±N+i(2N+1) where N is some positive integer.  As N goes to infinity through the integers, the integral vanishes on the left and right sides of the rectangle and along the top of the rectangle.  In particular, the absolute value of the integral along the top of the rectangle is bounded by  3N+1(2N+1)2−1∫−∞∞1coshπx2dx=6N+2(2N+1)2−1→0 as N  Then  ∫−∞∞x1+x21sinhπx2dx=2πi(Res[f(z),i]+∑k=1∞Res[f(z),2ki])  where  Res[f(z),i]=limz→izz+i1sinhπx2=12i  and  Res[f(z),2ki]=limz→2kiz2zsinhπz2+(1+z2)π2coshπz2  =4iπ(−1)kk1−4k2  And notice that  ∑k=1∞(−1)kk1−4k2=−14∑k=1∞((−1)k2k+1+(−1)k2k−1)

Daniel Cormack · Accepted Answer

I will solve the general form  ∫0∞xx2+1dxsinh(ax)=∫0∞xx2+a2dxsinh(x)  =∫0∞∫0∞e−atsin⁡(xt)sinh(x)dtdx  =∫0∞e−at∫0∞sin⁡(xt)sinh(x)dxdt  =π2∫0∞e−attanh(x)dt; z=2πa  =∫0∞e−zx(1−e−2x)e−2x+1dx  By spliting the integral we have  ∫0∞e−zxe−2x+1dx=∑n≥0∫0∞e−x(2n+z)dx  =∑n≥0(−1)n2n+z  =14(ψ(12+z4)−ψ(1+z4)−ψ(z4))  =12ψ(12+z4)−12ψ(z4)−z  =12ψ(12+a2π)−12ψ(a2

Vasquez · Accepted Answer

I≡∫0∞ln⁡(1+x2)cosh⁡(πx/2)sinh2⁡(πx/2)dx=2−4πI=−1π∫x→−∞x→∞ln⁡(1+x2)d[1sinh⁡(πx/2)]=2π∫−∞∞1sinh⁡(πx/2)xx2+1dx=2π[2πi∑n=1∞1πcosh⁡(nπi)/22ni(2ni)2+1+2πi1sinh⁡(πi/2)i2i]=16π∑n=1∞(−1)nn4n2−1+2=2+4π[∑n=1∞(−1)n2n−1+∑n=1∞(−1)n2n+1]=2+4π[−∑n=0∞(−1)n2n+1+∑n=1∞(−1)n2n+1]=2+4π[−1−∑n=1∞(−1)n2n+1+∑n=1∞(−1)n2n+1]=2−4π≈0.7268

Sovle integral: \int_0^\infty\log(1+x^2)\frac{\cosh\frac{\pi x}{2}}{\sinh^2\frac{\pi x}{2}}dx=2-\frac{4}{\pi}

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