Saturday, 19 September 2015

calculus - Show inti0nftyfraccosaxcosbxsinhbetaxfracdxx=logbig(fraccoshfracbpi2betacoshfracapi2betabig)



Hi I am trying to prove this interesting integral
I:=0cosaxcosbxsinhβxdxx=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=exex, but that didn't help either. Thanks


Answer



First I'm going to evaluate 0sintxsinhβxdx , (tR,Re(β)>0).



0sintxsinhβxdx=20sintxeβxeβxdx=20sintx eβx1e2βxdx=20sintxn=0e(2n+1)βxdx=2n=00sintx e(2n+1)βxdx=2n=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,bR,



0cosaxcosbxsinhβxdxx=0basinxtsinhβxdtdx=ba0sintxsinhβxdxdt=π2βbatanh(πt2β)dt=πb2βπa2βtanhudu=log(coshπb2β)log(coshπa2β)=log(coshπb2βcoshπa2β)


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