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 constant, 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 and the imaginary parts, the formulas $$\displaystyle\int_{-\infty}^\infty\frac{a\cos{x}}{x^2\!+\!a^2}\,dx \;=\; \int_{-\infty}^\infty\frac{x\sin{x}}{x^2\!+\!a^2}\,dx \;=\; \pi e^{-a}.$$

Bibliography

1

R. NEVANLINNA & V. PAATERO: Funktioteoria. Kustannusosakeyhtiö Otava. Helsinki (1963).