How do you integrate $\int_{0}^{\infty}\frac{a\cos{(cx)}}{a^2+x^2}dx$?

Question

This integral has been haunting me for a while now, as it's eluded every method of integration I could come up with (u-substitution, integration by parts, trigonometric substitution, and even Feynman's method). I realize it's non-elementary, but I can't figure out how to find the definite integral. I know you'll need to use Feynman's method, but I'm at a loss.

I don't know if this will help, but a and c are both positive.

To be clear, I want to know how to integrate it, not what the value is.

Answer 1

$$ \int_{0}^{\infty}\frac{a\cos{(cx)}}{a^2+x^2}\mathrm dx = ? $$

Advanced Calculus

This is the well-known Laplace integral.

WLOG, assume $a,c > 0$

Let the integral be $I (a,c)$, then

$$ \newcommand{\abs}[1]{\left\vert #1 \right\vert} \newcommand\rme{\mathrm e} \newcommand\imu{\mathrm i} \newcommand\diff{\,\mathrm d} \DeclareMathOperator\sgn{sgn} \renewcommand \epsilon \varepsilon \newcommand\trans{^{\mathsf T}} \newcommand\F {\mathbb F} \newcommand\Z{\mathbb Z} \newcommand\R{\Bbb R} \newcommand \N {\Bbb N} \newcommand\Q{\Bbb Q} \renewcommand \epsilon \varepsilon \DeclareMathOperator{\Ker}{Ker} \newcommand\bm\boldsymbol $$

$$ I (a, 0) = \int_0^{+\infty} \frac {a \diff x}{a^2 + x^2} = \frac \pi 2. $$

Taking the derivative, $$ \partial_c I = \int_0^{+\infty} \frac {-ax \sin (cx)}{x^2 + a^2}\diff x, $$ and by using the fact $$ \int_0^{+\infty} \frac {\sin (cx)}x \diff x = \frac \pi 2, $$ we get $$ \partial_c I + a\frac \pi 2 = a^2 \int_0^{+\infty} \frac {a\sin (cx)} {x (a^2 + x^2)}\diff x, $$ hence $$ \partial ^2_{cc} I = a^2\int_0^{+\infty} \frac {a\cos (cx)}{a^2 + x^2} = a^2 I(a,c). $$ Solve this ODE: the general solution is $$ I = C_1 \rme^{ac} + C_2 \rme ^{-ac}, $$ and since $$ \abs I \leqslant \int_0^{+\infty} \frac {a \diff x}{a^2 + x^2} = \frac \pi {2}, $$ $C_1$ shall be $0$, otherwise $\lim_{a \to +\infty} I = +\infty$, contradiction. Then according to $I(a, 0) = \pi /2$, we get $$ C_2 = \frac \pi 2, $$ then $$ \boxed {I (a,c) = \frac \pi 2 \rme^{-ac}}\ . $$

Complex Analysis

Consider $$ f(z) = \frac {\exp (\imu cz)}{z^2 + a^2} \quad [a >0, c>0], $$ and a contour $\gamma_R + I$ where $I$ is the interval $[-R, R]$ and $\gamma_R$ is the semicircle centering at $0$ with the radius $R$ that starts from $R + 0\imu$, where $R$ is large enough s.t. $R > a$. By the Cauchy integral theorem, $$ \int_{\gamma_R + I} f(z) \diff z = \int_{\abs {z - \imu a} = \varepsilon } f(z) \diff z = \int_{\abs {z -\imu a} } \frac {\dfrac {\exp (\imu cz)} {z+\imu a}} {z - \imu a} \diff z \stackrel ! = \imu 2\pi \cdot \frac {\exp (\imu c\cdot \imu a)}{2 \imu a} = \frac \pi a \rme ^{-ca}, (\bigstar) $$ where $!$ is the application of the Cauchy Integral Formula.

Now on $\gamma_R$, $z = R \rme^{\imu t}$ for $t \in [0, \pi]$, then $$ \abs {f(z)} = \abs {\frac {\rme^{\imu cz}}{a^2 + z^2} }= \abs {\frac {\exp (\imu c (R \cos t + \imu R \sin t))}{a^2 + R^2 \rme^{\imu 2t}} } = \frac {\exp (-cR \sin t)}{\abs {R^2 \rme^{\imu 2t} + a^2}} \leqslant \frac {\exp (-cR \sin t)}{R^2 - a^2} \leqslant \frac 1{R^2 - a^2} \xrightarrow {R \to +\infty} 0, $$ thus by taking the limit $R \to +\infty$ on the both side of $(\bigstar)$, $$ \boxed {\int_{-\infty}^{+\infty} \frac {\rme^{\imu cx}}{x^2 + a^2 } \diff x = \frac \pi a \rme ^{-ac} }\ . $$ Take the real part, we get $$ \int_{-\infty}^{+\infty} \frac {a \cos (cx)}{x^2 + a^2 } \diff x = \frac \pi 1 \rme ^{-ac}, $$ and since $\cos (\cdot)$ is even, we get $$ \boxed {I (a,c) = \frac \pi 2 \rme^{-ac}}\ . $$

Answer 2

This requires the use of the Laplace transform.

We set $$J(t;q)=q\int_0^\infty \frac{\cos(tx)dx}{x^2+q^2}.$$ Then we lake the Laplace transform of it: $$\begin{align} \mathcal{L}\{J(t;q)\}(s)&=\int_0^\infty e^{-st}J(t;q)dt\\ &=q\int_0^\infty \int_0^\infty \frac{e^{-st}\cos(tx)}{q^2+x^2}dxdt\\ &=q\int_0^\infty \frac{1}{x^2+q^2}\int_0^\infty e^{-st}\cos(tx)dtdx\\ &=qs\int_0^\infty \frac{dx}{(x^2+q^2)(x^2+s^2)}\\ &=\frac{qs}{q^2-s^2}\left[\int_0^\infty \frac{dx}{x^2+s^2}-\int_0^\infty \frac{dx}{x^2+q^2}\right]\\ &=\frac{\pi qs}{q^2-s^2}\left[\frac{1}{2s}-\frac{1}{2q}\right]\\ &=\frac{\pi}{2}\left[\frac{q}{q^2-s^2}-\frac{s}{q^2-s^2}\right]\\ &=\frac{\pi}{2}\left[\mathcal{L}\{\sinh qt\}(s)-\mathcal{L}\{\cosh qt\}(s)\right]\\ &=-\frac{\pi}{2}\mathcal{L}\{\cosh qt-\sinh qt\}(s)\\ &=-\frac{\pi}{2}\mathcal{L}\{e^{-qt}\}(s). \end{align}$$ Thus, when we take the inverse Laplace transform on both sides, we get $$J(t;q)=-\frac\pi2 e^{-qt}\ .$$