lemma, whose other hypotheses are satisfied. Define then a new function The pole has been moved away from the real axis, so can be integrated along the semicircle of radius centered at and closed on the real axis. One then takes the limit . The complex integral is zero by the residue theorem, as there are no poles inside the integration path The second term vanishes as goes to infinity. As for the first integral, one can use one version of the Sokhotski–Plemelj theorem for integrals over the real line: for a complex-valued function defined and continuously differentiable on the real line and real constants and with one finds where denotes the Cauchy principal value. Back to the above original calculation, one can write By taking the imaginary part on both sides and noting that the function is even, we get Finally, Alternatively, choose as the integration contour for the union of upper half-plane semicircles of radii and together with two segments of the real line that connect them. On one hand the contour integral is zero, independently of and ; on the other hand, as and the integral's imaginary part converges to (here is any branch of logarithm on upper half-plane), leading to . Dirichlet kernel Let be the Dirichlet kernel. It immediately follows that Define Clearly, is continuous when , to see its continuity at 0 apply L'Hopital's Rule: Hence, fulfills the requirements of the Riemann-Lebesgue Lemma. This means (The form of the Riemann-Lebesgue Lemma used here is proven in the article cited.) We would like to say that In order to do so, however, we must justify switching the real limit in to the integral limit in . This is in fact justified if we can show the limit does exist, which we do now. Using integration by parts, we have: Now, as and the term on the left converges with no problem. See the list of limits of trigonometric functions. We now show that is absolutely integrable, which implies that the limit exists. First, we seek to bound the integral near the origin. Using the Taylor-series expansion of the cosine about zero, Therefore, Splitting the integral into pieces, we have for some constant . This shows that the integral is absolutely integrable, which implies the original integral exists, and switching from to was in fact justified, and the proof is complete. See also Dirichlet distribution Dirichlet