# Laplace integrals

The improper integrals

 $\displaystyle\int_{-\infty}^{\infty}\frac{a\cos{x}}{x^{2}\!+\!a^{2}}\,dx\quad% \mbox{and}\quad\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^{2}\!+\!a^{2}}\,dx,$

where $a$ is a positive , are called Laplace integrals.  Both of them have the same value $\pi e^{-a}$.

The evaluation of the Laplace integrals can be performed by first determining the integrals

 $\int_{-\infty}^{\infty}\frac{e^{ix}}{x-ia}\,dx\quad\mbox{and}\quad\int_{-% \infty}^{\infty}\frac{e^{ix}}{x+ia}\,dx$

where one integrates along the real axis.  Therefore one has to determine the integrals

 $\oint\frac{e^{iz}}{z-ia}\,dz\quad\mbox{and}\quad\oint\frac{e^{iz}}{z+ia}\,dz$

around the perimeter of the half-disk with the arc in the upper half-plane, centered in the origin and with the diameter  $(-R,\,+R)$.  The residue theorem yields the values

 $\oint\frac{e^{iz}}{z-ia}\,dz\;=\;2i\pi e^{-a}\quad\mbox{and}\quad\oint\frac{e^% {iz}}{z+ia}\,dz\;=\,0.$

As in the entry example of using residue theorem, the parts of these contour integrals along the half-circle tend to zero when  $R\to\infty$.  Consequently,

 $\int_{-\infty}^{\infty}\frac{e^{ix}}{x-ia}\,dx\;=\;2i\pi e^{-a}\quad\mbox{and}% \quad\int_{-\infty}^{\infty}\frac{e^{ix}}{x+ia}\,dx\;=\;0.$

These equations imply by adding and subtracting and then taking the real (http://planetmath.org/RealPart) and the imaginary parts, the

 $\displaystyle\int_{-\infty}^{\infty}\frac{a\cos{x}}{x^{2}\!+\!a^{2}}\,dx\;=\;% \int_{-\infty}^{\infty}\frac{x\sin{x}}{x^{2}\!+\!a^{2}}\,dx\;=\;\pi e^{-a}.$

## References

• 1 R. Nevanlinna & V. Paatero: Funktioteoria.  Kustannusosakeyhtiö Otava. Helsinki (1963).
Title Laplace integrals LaplaceIntegrals 2013-03-22 18:43:17 2013-03-22 18:43:17 pahio (2872) pahio (2872) 5 pahio (2872) Definition msc 40A10