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

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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 $C \in \R$.


This needs considerable tedious hard slog to complete it.
In particular: This obviously only gives us an expression for $\ln \size {\map I b}$, we then need to determine that $\map I b > 0$ which seems nontrivial
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
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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.

$\blacksquare$


Sources

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