Using double integrals to solve single integrals. Issue 548 (14th June 2016) - Jameson, G. J. O.
Abstract : Consider the integral where b > a > 0. First, let us clarify why it even exists. Of course, convergence at infinity is ensured by the exponential terms, but the integrals of and e – ax / x and e – bx / x, taken separately, are divergent at 0, since these integrands equate asymptotically to 1/ x as x → 0. However, so ( e – ax – e – bx )/ x tends to the finite limit b – a as x → 0 and there is no problem integrating it on intervals of the form [0, r ]. A neat way to evaluate I 1 starts by expressing the integrand itself as an integral: (1) Inserting this into I 1 converts it into a double integral.