Hi I am trying to prove this interesting integral
I:=∫∞0cosax−cosbxsinhβxdxx=log(coshbπ2βcoshaπ2β),Re(β)>0.
I was thinking this was possibly a Frullani integral
∫∞0f(ax)−f(bx)xdx=[f(0)−f(∞)]logba,
but am unable to get rid of the sinhβx in the denominator. I have tried partial integration and splitting the integral up but ran into convergent issues. How can we solve this integral? Possibly we can go to exponential representation using 2sinhx=ex−e−x, but that didn't help either. Thanks
Answer
First I'm going to evaluate ∫∞0sintxsinhβxdx , (t∈R,Re(β)>0).
∫∞0sintxsinhβxdx=2∫∞0sintxeβx−e−βxdx=2∫∞0sintx e−βx1−e−2βxdx=2∫∞0sintx∞∑n=0e−(2n+1)βxdx=2∞∑n=0∫∞0sintx e−(2n+1)βxdx=2∞∑n=0t(2n+1)2β2+t2=π2βtanh(πt2β),
where I used the partial fraction expansion tanhz=∞∑n=02z((2n+1)π2)2+z2
Then for a,b∈R,
∫∞0cosax−cosbxsinhβxdxx=∫∞0∫basinxtsinhβxdtdx=∫ba∫∞0sintxsinhβxdxdt=π2β∫batanh(πt2β)dt=∫πb2βπa2βtanhudu=log(coshπb2β)−log(coshπa2β)=log(coshπb2βcoshπa2β)
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