Definite Integral to Infinity of Exponential of -a x^2 by Cosine of b x

Theorem

$\ds \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$

where $a$ is a strictly positive real number.

Proof

Fix $a$ and define:

$\ds \map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x$

for all $b \in \R$.

Then, we have:

 $\ds \map {I'} b$ $=$ $\ds \frac \d {\d b} \paren {\int_0^\infty e^{-a x^2} \cos b x \rd x}$ $\ds$ $=$ $\ds \int_0^\infty \frac \partial {\partial b} \paren {e^{-a x^2} \cos b x} \rd x$ Definite Integral of Partial Derivative $\ds$ $=$ $\ds -\int_0^\infty \paren {x e^{-a x^2} } \sin b x \rd x$ Derivative of $\cos a x$ $\ds$ $=$ $\ds -\paren {\intlimits {-\frac 1 {2 a} e^{-a x^2} \sin b x} 0 \infty - b \int_0^\infty \paren {-\frac 1 {2 a} e^{-a x^2} } \cos b x \rd x}$ Integration by Parts

Note that:

 $\ds \size {\frac 1 {2 a} e^{-a x^2} \sin b x}$ $\le$ $\ds \frac 1 {2 a} e^{-a x^2}$ noting that $\size {\sin x} \le 1$ $\ds$ $\to$ $\ds 0$ Exponential Tends to Zero and Infinity

So:

 $\ds -\paren {\intlimits {-\frac 1 {2 a} e^{-a x^2} \sin b x} 0 \infty - b \int_0^\infty \paren {-\frac 1 {2 a} e^{-a x^2} } \cos b x \rd x}$ $=$ $\ds -\frac b {2 a} \int_0^\infty e^{-a x^2} \cos b x \rd x$ $\ds$ $=$ $\ds -\frac b {2 a} \map I b$

We then have:

$\dfrac {\map {I'} b} {\map I b} = -\dfrac b {2 a}$

Integrating, by Primitive of Function under its Derivative and Primitive of Constant:

$\ln \size {\map I b} = -\dfrac {b^2} {4 a} + C$

for some $C \in \R$.

So:

$\map I b = A \map \exp {-\dfrac {b^2} {4 a} }$

for some $A \in \R$.

We have:

 $\ds \map I 0$ $=$ $\ds \int_0^\infty e^{-a x^2} \rd x$ $\ds$ $=$ $\ds \frac 1 2 \sqrt {\frac \pi a}$ Definite Integral to Infinity of $e^{-a x^2}$

on the other hand we have:

 $\ds \map I 0$ $=$ $\ds A \map \exp 0$ $\ds$ $=$ $\ds A$ Exponential of Zero

So we have:

$\ds\map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$

for all $b \in \R$ as required.

$\blacksquare$