# Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$

$$\displaystyle \int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx=\frac{\pi}{b}e^{-b\sqrt{a^2+c^2}}\sinh (ab)$$

I found this integral from here: http://integralsandseries.prophpbb.com/post2652.html?sid=d6641d4d4a3726f1b27bbb4b98ca840a and the solution uses contour integration. I am wondering if there is a way to solve it without using contour integration. I tried differentiating wrt $a$ and $c$ but in both cases, the resulting expression was dirty which made me reluctant to proceed further. I am out of ideas for this one.

Any help is appreciated. Thanks!

• Since the integrand has odd parity w.r.t. $b$, I experimented with the difference $I(a,b,c)-I(a,-b,c)=2I(a,b,c)=-\frac{2\pi}{b}\sinh{\left(b\sqrt{a^2+c^2}\right)}\sinh{(ab)}$, with the hope of exploiting hyperbolic trig identities for more insight into the symmetries of the problem. But then I realized that taking the sum of $I(b)$ and $I(-b)$ instead of their difference leads to a clear paradox. So I'm guessing there are probably unmentioned assumptions on the domains of the parameters that need to be specified for the internal consistency of the problem... Commented Jun 23, 2014 at 17:16
• @DavidH: Yes, that looks strange. The poster in the linked thread didn't mention anything about the domain so maybe we are already supposed to work within the domain where the result is true? But then, I am not sure what should be the correct domain, any ideas? Commented Jun 23, 2014 at 17:22
• Well, for each of the parameters $a,b,c$, if the domain includes $1$, then the domain also includes any positive real value because of scaling arguments (//waves hands vigorously//). So at the very least the domain includes the positive reals. Commented Jun 23, 2014 at 17:36
• Moving on to ideas for solving the integral itself, my gut tells me the methods for solving this integral math.stackexchange.com/questions/9402/… should be generalizable to the problem at hand. Commented Jun 23, 2014 at 18:43

In order to prove the final result I will need to state a lemma that will be used later.

Lemma$\require{autoload-all}$ $1$:

$$\int_0^\infty \! \frac{\cos(bx)}{x^2+\alpha} \mathrm{d}x = \frac{\pi e^{-b\sqrt{\alpha}}}{2b\sqrt{\alpha}}\tag{1}$$

Proof here.

Consider

$$I = \int_0^\infty\!\! \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; \mathrm{d}x$$

Integrate by parts

$$I = \int_0^\infty \!\!\frac{2 a \left(c^2-x^2\right) \cos (b x)}{x^4 +(4 a^2+2 c^2) x^2+c^4} \; \mathrm{d}x$$

Decompose this function by partial fractions $$\frac{2 a \left(c^2-x^2\right) }{x^4 +(4 a^2+2 c^2) x^2+c^4} = \frac{a_-}{x^2+x_0} + \frac{a_+}{x^2+x_1}$$

It so happens that

$$x_0 = 2 a^2+2 a\sqrt{a^2+c^2}+c^2,\quad x_1 = 2 a^2-2a \sqrt{a^2+ c^2}+c^2$$

$$a_- =\frac{2a(c^2+x_0)}{x_1-x_0}, \quad a_+ = \frac{2a(c^2+x_1)}{x_0-x_1}$$

Note that both $x_0$ and $x_1$ are greater than $0$.

Re-write the integral

\begin{align} I &= a_-\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x + a_+\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\\[.3cm] &= \frac{2a(c^2+x_0)}{x_1-x_0}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x +\frac{2a(c^2+x_1)}{x_0-x_1}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\end{align}

Using $(1)$:

\begin{align} I &= \frac{2a(c^2+x_0)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_0}}}{2b\sqrt{x_0}} - \frac{2a(c^2+x_1)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_1}}}{2b\sqrt{x_1}}\\[.3cm] &= \left(\frac{a\pi}{b(x_1-x_0)}\right)\left(\frac{(c^2+x_0)e^{-\sqrt{x_0}}}{\sqrt{x_0}} - \frac{(c^2+x_1)e^{-\sqrt{x_1}}}{\sqrt{x_1}}\right)\end{align}

I will digress here to state (without proof but easily verified) that $$\frac{c^2+x_0}{\sqrt{x_0}}= \frac{c^2+x_1}{\sqrt{x_1}} = \frac{x_1-x_0}{2a}.$$ This allows us a tremendous simplification so that we can write

\begin{align} I = \left(\frac{\pi}{2b}\right)\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right). \end{align}

It can also be shown that $$\sqrt{x_1} = -a+\sqrt{a^2+c^2}$$ $$\sqrt{x_0} = a+\sqrt{a^2+c^2}$$

Simply square each side to find the desired equality.

We can now complete the proof:

\begin{align} I &= \frac{\pi}{2b}\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right) \\[.2cm] &= \frac{\pi}{2b}\left(\exp\left(ab-b\sqrt{a^2+c^2}\right)-\exp\left(-ab-b\sqrt{a^2+c^2}\right) \right) \\[.2cm] &= \frac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\frac12(\exp(ab)-\exp(ab)) \\[.2cm] &= \dfrac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\sinh{ab}\end{align}

If you are interested in working through the simplifications that I did not prove, I recommend that you begin by squaring each side after verifying that each side shares the same sign.

• I think partial fractions is the only way. Can you please complete your solution so that I can check it off as the answer or shall I write down an answer? Thank you! :) Commented Jun 25, 2014 at 5:25
• I will finish it tonight when I am more available.
Commented Jun 25, 2014 at 13:50
• @PranavArora Just a ping to say that I have completed my answer. Feel free to ask if you need anything clarified.
Commented Jun 27, 2014 at 7:02
• Excellent work! Thanks a lot. :) Commented Jun 27, 2014 at 8:06

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x ={\pi \over \verts{b}} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}: \ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x} =\half\,\sgn\pars{ab}\Im\int_{-\infty}^{\infty} \arctan\pars{2\verts{a}x \over x^{2} + c^{2}}\expo{\ic\verts{b}x}\,\dd x \\[3mm]&=\half\,\sgn\pars{ab}\,\Im\int_{-\infty}^{\infty}{\ic \over 2}\,\ln\pars{% 1 - 2\verts{a}x\ic/\bracks{x^{2} + c^{2}}\over 1 + 2\verts{a}x\ic/\bracks{x^{2} + c^{2}}}\expo{\ic\verts{b}x}\,\dd x \\[3mm]&={1 \over 4}\,\sgn\pars{ab}\,\Re\int_{-\infty}^{\infty}\ln\pars{% x^{2}- 2\verts{a}\ic x + c^{2} \over x^{2} + 2\verts{a}\ic x + c^{2}} \ \underbrace{\expo{\ic\verts{b}x}\,\dd x} _{\ds{\dd\pars{\expo{\ic\verts{b}x} \over \ic\verts{b}}}} \end{align} $$\begin{array}{|c|}\hline \\ \quad\mbox{Here, we used the identity}\quad \arctan\pars{x} = {\ic \over 2}\,\ln\pars{1 - x\ic \over 1 + x\ic}\quad \\ \\ \hline \end{array}$$

Integrating by parts: \begin{align}&\color{#c00000}{% \int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x} \\[3mm]&=-\,{1 \over 4}\,\sgn\pars{ab}\,\Re\int_{-\infty}^{\infty} \pars{{2x - 2\verts{a}\ic \over x^{2} - 2\verts{a}\ic x + c^{2}}- {2x + 2\verts{a}\ic \over x^{2} + 2\verts{a}\ic x + c^{2}}} \expo{\ic\verts{b}x}\,{\dd x \over \ic\verts{b}} \\[3mm]&=-\,{\sgn\pars{a} \over 2b}\,\Im\int_{-\infty}^{\infty} \pars{{x - \verts{a}\ic \over x^{2} - 2\verts{a}\ic x + c^{2}}- {x + \verts{a}\ic\over x^{2} + 2\verts{a}\ic x + c^{2}}} \expo{\ic\verts{b}x}\,\dd x \end{align}

\begin{align} &\mbox{Zeros of}\quad x^{2} - 2\verts{a}\ic x + c^{2} =0 \quad\mbox{are given by}\quad \left\lbrace\begin{array}{rcl} \phantom{-}x_{1} & = &\pars{\verts{a} + \root{a^{2} + c^{2}}}\ic \\[2mm] \phantom{-}x_{2} & = & \pars{\verts{a} - \root{a^{2} + c^{2}}}\ic \end{array}\right. \\[3mm]&\mbox{Zeros of}\quad x^{2} + 2\verts{a}\ic x + c^{2} =0 \quad\mbox{are given by}\quad \left\lbrace\begin{array}{rcl} -x_{1} & = &\pars{-\verts{a} - \root{a^{2} + c^{2}}}\ic \\[2mm] -x_{2} & = & \pars{-\verts{a} + \root{a^{2} + c^{2}}}\ic \end{array}\right. \end{align} Note that $\ds{\Im\pars{x_{1}} > 0}$ and $\ds{\Im\pars{x_{2}} < 0}$.

Therefore, \begin{align} &\color{#c00000}{% \int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x} \\[3mm]&=-\,{\sgn\pars{a} \over 2b}\,\Im\pars{% 2\pi\ic\,{\pars{x_{1} - \verts{a}\ic}\expo{\ic\verts{b}x_{1}}\over 2x_{1} - 2\verts{a}\ic} - 2\pi\ic\,{\pars{-x_{2} + \verts{a}\ic}\expo{-\ic\verts{b}x_{2}}\over -2x_{2} + 2\verts{a}\ic}} \\[3mm]&=-\,{\pi\sgn\pars{a} \over 2b}\braces{% \exp\pars{-\verts{b}\bracks{\verts{a} + \root{a^{2} + c^{2}}}} - \exp\pars{\verts{b}\bracks{\verts{a} - \root{a^{2} + c^{2}}}}} \\[3mm]&=-\,{\pi\sgn\pars{a} \over 2b}\bracks{% \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}} \pars{\expo{-\verts{ab}} - \expo{\verts{ab}}}} \\[3mm]&={\pi\sgn\pars{a} \over b} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{\verts{ab}} ={\pi \over \verts{b}} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab} \end{align}

\begin{align} &\color{#66f}{\large% \int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x ={\pi \over \verts{b}} \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}} \end{align}

• Thank you Sir Felix but I forgot to state that I am looking for elementary methods. +1. :) Commented Jun 25, 2014 at 5:26
• @PranavArora I think the $\large\sin\left(bx\right)$ factor avoids we use any other methods. However, it's still interesting your point of view and I believe we have to think a little bit more about that. Thanks. Commented Jun 25, 2014 at 5:34
• Brad's solution might work, I am working on it to see if it gives the result. Commented Jun 25, 2014 at 5:35

When we see an arctan, sometimes parts is a good start.

Let $$\displaystyle u=\tan^{-1}\left(\frac{2ax}{x^{2}+c^{2}}\right), \;\ dv=\sinh(bx)dx, \;\ du=\frac{-2a(x+c)(x-c)}{x^{4}+2(2a^{2}+c^{2})x^{2}+c^{4}}dx, \;\ v=\frac{-1}{b}\cos(bx)$$

The uv part goes to 0 and we have remaining due to it being an even function:

$$\frac{2a}{b}\int_{-\infty}^{\infty}\frac{(x-c)(x+c)\cos(bx)}{x^{4}+2(2a^{2}+c^{2})x^{2}+c^{4}}dx$$

Use a semicircle in the UHP and consider:

$$f=\frac{2a}{b}\int_{C}\frac{(z+c)(z-c)e^{ibz}}{z^{4}+2(2a^{2}+c^{2})z^{2}+c^{4}}$$

The portion around the arc tends to 0 as $$R\to \infty$$.

It has poles at:

$$z=(\sqrt{a^{2}+c^{2}}+a)i, \;\ z=(\sqrt{a^{2}+c^{2}}-a)i$$

$$2\pi i \cdot Res(f, \;\ (\sqrt{a^{2}+c^{2}}+a)i)=\frac{\pi }{b}e^{-ab}e^{-b\sqrt{a^{2}+c^{2}}}$$

$$2\pi i \cdot Res(f, \;\ (\sqrt{a^{2}+c^{2}}-a)i)=\frac{\pi}{b}e^{ab}e^{-b\sqrt{a^{2}+c^{2}}}$$

sum the residues in the UHP and note the exponential identity for sinh(ab). Thus, obtaining:

$$=\frac{\pi}{b}e^{-b\sqrt{a^{2}+c^{2}}}\sinh(ab)$$