- Evaluate the integrals
\[ \int_ {0} ^ {\infty} e ^ {- a x} \cos b x d x \quad \text { and } \quad \int_ {0} ^ {\infty} e ^ {- a x} \sin b x d x, \quad a > 0 \]
by integrating \(e ^ { - A z } , A = { \sqrt { a ^ { 2 } + b ^ { 2 } } }\) , over an appropriate sector with angle \(\omega ,\) with cos \(\omega = a / A\)
Solution
Let \[ A=\sqrt{a^2+b^2}, \qquad \cos\omega=\frac aA, \qquad \sin\omega=\frac bA, \] with \(|\omega|<\pi/2\) because \(a>0\). Then \[ Ae^{-i\omega}=a-ib. \] Integrate the entire function \(e^{-Az}\) around the sector bounded by the positive real axis, the ray \(\arg z=-\omega\), and a circular arc of radius \(R\) joining them. On the arc, \(z=Re^{-it}\) with \(t\) between \(0\) and \(\omega\), so \(\Re(Az)=AR\cos t\ge AR\cos\omega=aR\). Hence \(\abs{e^{-Az}}\le e^{-aR}\) there, and the arc integral is at most \(R\abs\omega e^{-aR}\to0\) as \(R\to\infty\).
The two radial sides therefore give \[ \int_0^\infty e^{-Ax}\,dx =e^{-i\omega}\int_0^\infty e^{-Ae^{-i\omega}x}\,dx. \] Since the left side is \(1/A\), \[ \int_0^\infty e^{-(a-ib)x}\,dx =\frac{e^{i\omega}}A =\frac{a+ib}{a^2+b^2}. \] But \[ e^{-(a-ib)x}=e^{-ax}(\cos bx+i\sin bx). \] Taking real and imaginary parts yields \[ \boxed{\displaystyle \int_0^\infty e^{-ax}\cos bx\,dx=\frac{a}{a^2+b^2}}, \qquad \boxed{\displaystyle \int_0^\infty e^{-ax}\sin bx\,dx=\frac{b}{a^2+b^2}}. \]