Laplace integrals

The improper integrals

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{a\cos x}{x^2+a^2}𝑑x\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x\sin x}{x^2+a^2}𝑑x,$$

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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{x-ia}𝑑x\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{x+ia}𝑑x$$

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

$$\oint \frac{e^{iz}}{z-ia}𝑑z\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\oint \frac{e^{iz}}{z+ia}𝑑z$$

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

$$\oint \frac{e^{iz}}{z-ia}𝑑z=\mathrm{\hspace{0.33em}2}i\pi e^{-a}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\oint \frac{e^{iz}}{z+ia}𝑑z=\mathrm{\hspace{0.17em}0}.$$

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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{x-ia}𝑑x=\mathrm{\hspace{0.33em}2}i\pi e^{-a}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{x+ia}𝑑x=\mathrm{\hspace{0.33em}0}.$$

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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{a\cos x}{x^2+a^2}𝑑x={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x\sin x}{x^2+a^2}𝑑x=\pi e^{-a}.$$