# Prove that $\int_0^\infty \frac{e^{\cos(ax)}\cos\left(\sin (ax)+bx\right)}{c^2+x^2}dx =\frac{\pi}{2c}\exp\left(e^{-ac}-bc\right)$

In my course, I have to prove formula below $$I=\int_0^\infty \frac{e^{\cos(ax)}\cos\left(\sin (ax)+bx\right)}{c^2+x^2}dx =\frac{\pi}{2c}\exp\left(e^{-ac}-bc\right)$$ for $a,b,c>0.$

I know that this integral can be easily solved with complex analysis using $$f(z)=\frac{1}{2} \ \mathbb{R} \left(\int_{-\infty}^\infty \frac{\exp\left(e^{iaz}+ibz\right)}{c^2+z^2}dz\right)$$ but right now I am in a course dealing with real analysis. I tried to use parametrization integral method $$I'(a)=-\int_0^\infty \frac{xe^{\cos(ax)}\sin(\sin(ax)+(a+b)x)}{c^2+x^2}dx$$ but it doesn't look easier to handle. I tried to differentiate it again, but I just got a horrible form. An idea came to mind to differentiate with respect to parameter $b$ and set a differential equation $$I''(b)+x^2I(b)=0$$ plugging this ODE to W|A, I got $$I(b)=c_1\cos(bx^2)+c_2\sin(bx^2)$$ It's definitely wrong! After seeing Samrat's answer, I tried to plug in again to W|A and I got $$I(b)=c_1 D_{-1/2}((i+1)b)+c_2 D_{-1/2}((i-1)b)$$

where $D_n(z)$ is the parabolic cylinder function but I have no idea what does that mean.

• The result seems off. What if $b \lt 0$? – Ron Gordon Nov 5 '14 at 15:59
• Sorry, I forgot to mention the formula holds for $a,b,c>0$ – Venus Nov 5 '14 at 16:00
• Thanks. That's a nice detail to know up front. – Ron Gordon Nov 5 '14 at 16:11

\begin{aligned} \int_0^{\infty} \frac{e^{\cos(ax)}\cos(\sin(ax)+bx)}{x^2+c^2}\,dx &=\Re\left(\int_0^{\infty} \frac{e^{e^{iax}}e^{ibx}}{x^2+c^2}\right) \\ &=\Re\left(\sum_{k=0}^{\infty} \frac{1}{k!}\int_0^{\infty} \frac{e^{i(ak+b)x}}{x^2+c^2}\,dx\right)\\ &=\sum_{k=0}^{\infty} \frac{1}{k!}\int_0^{\infty}\frac{\cos((ak+b)x)}{x^2+c^2}\,dx \\ &=\frac{\pi}{2c}\sum_{k=0}^{\infty} \frac{1}{k!}e^{-c(ak+b)}\\ &=\frac{\pi}{2c}e^{-bc}\sum_{k=0}^{\infty} \frac{e^{-kac}}{k!}\\ &=\frac{\pi}{2c}e^{-bc}e^{e^{-ac}}=\boxed{\dfrac{\pi}{2c}\exp\left(e^{-ac}-bc\right)} \\ \end{aligned}

I used the following result: $$\int_0^{\infty} \frac{\cos(mx)}{x^2+a^2}=\frac{\pi}{2a}e^{-am}$$

• Wow! It's very simple. Thanks Pranav! You help me again :-) – Venus Nov 6 '14 at 9:22
• @Venus: Glad to help! :) – Pranav Arora Nov 6 '14 at 16:14
• What is the fancy R? – Joao Nov 7 '14 at 1:26

As a generalization of Pranav's answer, let us assume that $f(z)$ has a Maclaurin series expansion with real coefficients that converges absolutely on the unit circle on the complex plane.

Then for $a, b, c >0$,

\begin{align} \text{Re} \int_{0}^{\infty} \frac{e^{ibx} f(e^{i a x}) }{c^{2}+x^{2}} \ dx &= \text{Re} \int_{0}^{\infty} \frac{e^{ibx}}{c^{2}+x^{2}} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} e^{ianx}\ dx \\ &= \text{Re} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{e^{i(an+b)x}}{c^{2}+x^{2}} \ dx \tag{1} \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\cos[(an+b)x]}{c^{2}+x^{2}} \ dx \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \frac{\pi}{2c} e^{-c(an+b)} \\ &= \frac{\pi}{2c} e^{-bc} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} e^{-acn} \\ &= \frac{\pi}{2c} e^{-bc} f(e^{-ac}). \end{align}

Your integral is the case $f(z) = e^{z}$.



• Thanks for your answer, +1. I learned much from your posts here. Thank you so much! :-) – Venus Nov 7 '14 at 10:36
• @Venus You're very welcome. – Random Variable Nov 7 '14 at 11:37
• RV What did you mean by "the Taylor expansion converges absolutely for values on the unit circle"? Could you give me examples of $f(z)$ that satisfy the given condition? – Venus Nov 25 '14 at 11:13
• @Venus That the radius of convergence of the Taylor series is exactly $1$ or greater. Otherwise I wouldn't be able to replace the complex variable $z$ by $e^{iax}, x \in \mathbb{R}$. Functions that satisfy this condition include $e^{z}, \cos(z), \sin(z)$, and $z^{n}$ where $n$ is a positive integer. An example of a function that doesn't is $\frac{1}{\sqrt{1-4z}}$. – Random Variable Nov 25 '14 at 12:06
• How about $\displaystyle\frac{1}{1\pm z}$ or $\displaystyle\frac{e^z}{z}$? – Venus Nov 25 '14 at 12:37

Note that $$I''(b)=-\int_{0}^{\infty}\frac{x^2e^{\cos ax}\cos(\sin (ax)+bx) }{c^2+x^2}dx\ne -x^2I(b)$$ That is your mistake, otherwise, the approach is fine.

• I did say this was wrong in my question – Venus Nov 5 '14 at 15:56
• Yeah, I was just showing you where you made the mistake. Your idea is not wrong, but taking out $x^2$ is wrong. If you proceed now, I think you will be able to get the answer. – Samrat Mukhopadhyay Nov 5 '14 at 15:57
• I got this wolframalpha.com/input/?i=y%27%27+%2Bx^2+y%3D0 – Venus Nov 5 '14 at 15:58