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

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.

• – nmasanta Jun 14 '19 at 1:36
• The usual way to do this kind of thing involves integrals in the complex plane. Have you studied that yet? – David Jun 14 '19 at 1:39
• @David - Ah, no I have not. I know about the complex plane, but I haven't done calculus on it yet. I would appreciate an answer, but it's quickly becoming apparent that this integral is beyond my relatively small knowledge of mathematics. – The_Scientist___ Jun 14 '19 at 1:59

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

This is the well-known Laplace integral.

WLOG, assume $$a,c > 0$$

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


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

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 $$\boxed {I (a,c) = \frac \pi 2 \rme^{-ac}}\ .$$

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 $$J(t;q)=-\frac\pi2 e^{-qt}\ .$$

• I would only that that you needed to appeal to Fubini's Theorem in evaluating the Laplace Transform. Great Solution btw (+1) – user679268 Jul 1 '19 at 10:47
• Sorry that and the Dominated Convergence Theorem. – user679268 Jul 1 '19 at 10:48