## Monthly Archives: January 2017

### Another integral that Mathematica cannot do

The integral to evaluate is $$\int_0^{\infty} dx \frac{\sin{\left (\pi x^2 \right )}}{\sinh^2{(\pi x)}} \cosh{(\pi x)}$$ Given the trig functions in the integrand, it makes sense to use the residue theorem based on a complex integral around a rectangular contour. As has been my experience with these integrals, the integrand of the complex integral will not […]