Definite Integral to Infinity of Cosine m x over x Squared plus a Squared

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Theorem

$\ds \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi {2 a} e^{-m a}$

where $m$ and $a$ are positive real numbers.


Proof

From Definite Integral of Even Function:

$\ds \frac 1 2 \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x$

Let $R$ be a positive real number with $R > a$.

Let $C_1$ be the straight line segment from $-R$ to $R$.

Let $C_2$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $-R$ counter-clockwise.

Let $\Gamma = C_1 \cup C_2$.

Let:

$\map f z = \dfrac {e^{i m z} } {z^2 + a^2}$

From Euler's Formula, we have:

$\map f z = \dfrac {\cos m z} {z^2 + a^2} + i \dfrac {\sin m z} {z^2 + a^2}$

So:

\(\ds \int_{-\infty}^\infty \map f x \rd x\) \(=\) \(\ds \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x + i \int_{-\infty}^\infty \frac {\sin m x} {x^2 + a^2} \rd x\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x\) Definite Integral of Odd Function

Note that the integrand is meromorphic with simple poles where $z^2 + a^2 = 0$.

That is, at $z = a i$ and $z = -a i$.

As our semi-circular contour lies in the upper half-plane, the only pole of concern is $a i$.

As $R > a$, these poles do not lie on $C_2$, but are enclosed by the curve $\Gamma$.

We have:

\(\ds \int_C \map f z \rd z\) \(=\) \(\ds \int_{C_1} \frac {e^{i m z} } {z^2 + a^2} \rd z + \int_{C_2} \frac {e^{i m z} } {z^2 + a^2} \rd z\) Contour Integral of Concatenation of Contours
\(\ds \) \(=\) \(\ds \int_{-R}^R \frac {e^{i m x} } {x^2 + a^2} \rd x + \int_{C_2} \frac {e^{i m z} } {z^2 + a^2} \rd z\) Definition of Contour Integral

The integral over $C_2$ can be shown to vanish as $R \to \infty$:

\(\ds \size {\int_{C_2} \frac {e^{i m z} } {z^2 + a^2} \rd z}\) \(\le\) \(\ds \frac \pi m \max_{0 \le \theta \le \pi} \size {\frac 1 {R^2 e^{2 i \theta} + 1} }\) Jordan's Lemma
\(\ds \) \(=\) \(\ds \frac \pi m \paren {\frac 1 {R^2 - 1} }\)
\(\ds \) \(\to\) \(\ds 0\) as $R \to \infty$

Taking $R \to \infty$, we have:

\(\ds \int_{-\infty}^\infty \frac {e^{i m x} } {x^2 + a^2} \rd x\) \(=\) \(\ds \int_C \frac {e^{i m z} } {z^2 + a^2} \rd z\)
\(\ds \) \(=\) \(\ds 2 \pi i \Res {\frac {e^{i m z} } {z^2 + a^2} } {a i}\) Residue Theorem
\(\ds \) \(=\) \(\ds 2 \pi i \paren {\frac {e^{i m z} } {2 z} }_{z = a i}\) Residue at Simple Pole
\(\ds \) \(=\) \(\ds 2 \pi i \frac {e^{-m a} } {2 a i}\)
\(\ds \) \(=\) \(\ds \frac \pi a e^{-m a}\)

So:

$\ds \int_{-\infty}^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi a e^{-m a}$

giving:

$\ds \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi {2 a} e^{-m a}$

$\blacksquare$


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