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2008-05-18T19:11:21Z

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Vipul: New page: ==Description== Here, we describe the integration procedure for integrals of the form: $\operatorname{PV} \int_{-\infty}^\infty \frac{P(x) \sin x}{Q(x)}$ The $\sin$
2008-04-29T00:16:20Z

New page: ==Description== Here, we describe the integration procedure for integrals of the form: <math>\operatorname{PV} \int_{-\infty}^\infty \frac{P(x) \sin x}{Q(x)}</math> The <math>\sin</math...

New page
==Description==

Here, we describe the integration procedure for integrals of the form:

<math>\operatorname{PV} \int_{-\infty}^\infty \frac{P(x) \sin x}{Q(x)}</math>

The <math>\sin</math> may be replaced by <math>\cos</math>.

Here, we assume that the polynomial <math>Q</math> does not have any zeros on the real axis (except possibly at the origin).

==Solution by semicircular contour integration==

{{further|[[semicircular contour integration method]]}}

This is typical, for instance, when the integrand is of the form:

<math>f(x) := \frac{\sin x}{Q(x)}</math>

where the degree of <math>Q</math> is strictly greater than one, and <math>Q</math> has no real zeros. We use the complex-valued function <math>e^{iz}/Q(z)</math>. In this case, [[Jordan's lemma]] (or the weaker version, the so-called [semicircular contour theorem]]) tells us that the integral along large semicircles tends to zero, so:

<math>\operatorname{PV} \int_{-\infty}^\infty \frac{\sin x}{Q(x)} = \operatorname{Im} \left((2 \pi i) \sum \operatorname{res} (f;z_j)\right)</math>

where the sum is taken over all the zeros <math>z_j</math> of <math>f</math> that lie in the upper half-plane.

Note that we can drop the <math>\operatorname{PV}</math> ([[Cauchy principal value]] symbol), since the integral is absolutely convergent in this case. Also,if <math>Q</math> has only simple poles, so that we have:

<math>Q(z) := \prod_{j=1}^n (z - z_j)</math>

Then, we get the formula:

<math>\int_{-\infty}^\infty \frac{\sin x}{Q(x)} = 2 \pi \operatorname{Re} \left(\sum_{j=1}^n \frac{\sin(z_j)}{\prod_{k \ne j} (z_j - z_k)}\right)</math>