Often, the half-circle part of the integral will tend towards zero as the radius of the half-circle grows, leaving only the real-axis part of the integral, the one we were originally interested in. Examples An integral along the real axis The integral arises in probability theory when calculating the characteristic function of the Cauchy distribution. It resists the techniques of elementary calculus but can be evaluated by expressing it as a limit of contour integrals. Suppose and define the contour that goes along the real line from to and then counterclockwise along a semicircle centered at 0 from to . Take to be greater than 1, so that the imaginary unit is enclosed within the curve. Now consider the contour integral Since is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator is zero. Since , that happens only where or . Only one of those points is in the region bounded by this contour. Because is the residue of at is According to the residue theorem, then, we have The contour may be split into a straight part and a curved arc, so that and thus Using some estimations, we have and The estimate on the numerator follows since , and for complex numbers along the arc (which lies in the upper half-plane), the argument of lies between 0 and . So, Therefore, If then a similar argument with an arc that winds around rather than shows that and finally we have (If then the integral yields immediately to elementary calculus methods and its value is .) An infinite sum The fact that has simple poles with residue 1 at each integer can be used to compute the sum Consider, for example, . Let be the rectangle that is the boundary of with positive orientation, with an integer . By the residue formula, The left-hand side goes to zero as since the integrand has order . On the other hand, where the Bernoulli number (In fact, .) Thus, the residue is . We conclude: which is a proof of the Basel problem. The same trick can be used to establish the sum of the Eisenstein series: We take with a non-integer and we shall show the above for . The difficulty in this case is to show the vanishing of the contour integral at infinity. We have: