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

Question

Please help me in proving the following result:

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

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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!

Answer 1

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

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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.

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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 $0$.

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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 $(1)$:

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

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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.

Answer 2

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

\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}

Answer 3

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:

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