# Integrate sin(bx)/sinh(ax) from zero to infinity

Do the integral (0 --> infinity) using contour integration:

J(a,b) = dx sin(bx)/sinh(ax)

https://brainmass.com/math/integrals/integrate-sin-bx-sinh-ax-zero-infinity-530903

#### Solution Preview

We note that the integrand is an even function of x, so we can write:

J(a,b) = 1/2 Integral from minus to plus infinity of sin(bx)/sinh(ax) dx

It is convenent to scale out eiter of the constants a or b. Let's get rid of a by putting ax = u:

J(a,b) = 1/(2a) Integral from minus to plus infinity of sin(b/a u)/sinh(u) du

If we put k = b/a, then we can write:

Integral from minus to plus infinity of sin(k x)/sinh(x) dx =

Integral from minus to plus infinity of Im[exp(i k x)]/sinh(x) dx =

Limit epsilon to zero of

Integral from minus infinity to minus epsilon and then from epsilon to plus infinity of Im[exp(i k x)]/sinh(x) dx =

Limit epsilon to zero of

Im [ Integral from minus infinity to minus epsilon and then from epsilon to plus infinity of exp(i k x)/sinh(x) dx ] =

Limit epsilon to zero and R to infinity of

Im [ Integral from minus R to minus epsilon and then from epsilon to R of exp(i k x)/sinh(x) dx ] (1)

To evaluate this, we first assume k that k>=0 and consider the contour integral of exp(i k z)/sinh(z) from -R to -epsilon, and then along a counterclockwise semi-circle of radius epsilon with center the origin, we travel from minus epsilon to epsilon, and then from epsilon we go to R, and then along a counterclockwise semi-circle of radius R with center the origin, we travel back from R to minus R.

Then this contour integral includes ...

#### Solution Summary

We explain in detail using the methods of complex analysis, how to evaluate the integral.