Definite Integral to Infinity of x by Sine m x over x Squared plus a Squared
Theorem
- $\ds \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \frac \pi 2 e^{-m a}$
where $m$ and $a$ are positive real numbers.
Proof 1
From Definite Integral of Even Function:
- $\ds \frac 1 2 \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x$
With the aim of integrating over the domain, we split the domain up into $2$ components as follows:
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 $-R$ anticlockwise.
Let $\Gamma = C_1 \cup C_2$.
Let:
- $\map f x = \dfrac {x e^{i m x} } {x^2 + a^2}$
From Euler's Formula, we have:
- $\map f x = \dfrac {x \cos m x} {x^2 + a^2} + i \dfrac {x \sin m x} {x^2 + a^2}$
So:
\(\ds \int_{-\infty}^\infty \map f x \rd x\) | \(=\) | \(\ds \int_{-\infty}^\infty \frac {x \cos m x} {x^2 + a^2} \rd x + i \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x\) | Linear Combination of Definite Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds i \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x\) | Definite Integral of Odd Function |
Note that the integrand is meromorphic with simple poles where $x^2 + a^2 = 0$.
That is, at $x = a i$ and $x = -a i$.
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:
\(\ds \int_{\Gamma} \map f x \rd x\) | \(=\) | \(\ds \int_{C_1} \frac {x e^{i m x} } {x^2 + a^2} \rd x + \int_{C_2} \frac {x e^{i m x} } {x^2 + a^2} \rd x\) | Contour Integral of Concatenation of Contours | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{-R}^R \frac {x e^{i m x} } {x^2 + a^2} \rd x + \int_{C_2} \frac {x e^{i m x} } {x^2 + a^2} \rd x\) | Definition of Complex Contour Integral |
The integral over $C_2$ can be shown to vanish as $R \to \infty$:
\(\ds \size {\int_{C_2} \frac {x e^{i m x} } {x^2 + a^2} \rd x}\) | \(\le\) | \(\ds \frac \pi m \max_{0 \le \theta \le \pi} \size {\frac {R e^{i \theta} } {R^2 e^{2 i \theta} + 1} }\) | Jordan's Lemma | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi m \paren {\frac R {R^2 - 1} }\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | as $R \to \infty$ |
Taking $R \to \infty$, we have:
\(\ds \int_{-\infty}^\infty \frac {x e^{i m x} } {x^2 + a^2} \rd x\) | \(=\) | \(\ds \int_{C_1} \frac {x e^{i m x} } {x^2 + a^2} \rd x + 0\) | The integral over $C_2$ vanished | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi i \Res {\frac {x e^{i m x} } {x^2 + a^2} } {a i}\) | Cauchy's Residue Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi i \paren {\lim_{x \mathop \to ai} \paren {x - a i} \frac {x e^{i m x} } {\paren {x + a i} \paren {x - a i} } }\) | Residue at Simple Pole | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi i \paren {\frac {x e^{i m x} } {x + a i} }_{x = a i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi i e^{-m a}\) |
So:
- $\ds \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \pi e^{-m a}$
giving:
- $\ds \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \frac \pi 2 e^{-m a}$
$\blacksquare$
Proof 2
From Definite Integral to Infinity of $\dfrac {\cos m x} {x^2 + a^2}$:
- $\ds \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi {2 a} e^{-m a}$
We have:
\(\ds \frac \d {\d m} \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x\) | \(=\) | \(\ds \int_0^\infty \frac \partial {\partial m} \paren {\frac {\cos m x} {x^2 + a^2} } \rd x\) | Definite Integral of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds -\int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x\) | Derivative of $\cos a x$ |
So:
\(\ds \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x\) | \(=\) | \(\ds -\frac \d {\d m} \paren {\frac \pi {2 a} e^{-m a} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 2 e^{-m a}\) | Derivative of $e^{a x}$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.41$