Integral $\int_{-\infty}^{\infty}\frac{e^{r \arctan(ax)}+e^{-r \arctan(ax)}}{1+x^2}\cos \left( \frac{r}{2}\log(1+a^2x^2)\right)dx$

How can I show that

$$\int_{-\infty}^{\infty}\frac{e^{r \arctan(ax)}+e^{-r \arctan(ax)}}{1+x^2}\cos \left( \frac{r}{2}\log(1+a^2x^2)\right)dx = 2\pi \cos \left( r\log(1+a)\right)$$ where $a \in \mathbb{R}^+$ and $r \in \mathbb{R}$?

The relatively simple result suggests that it might be possible to evaluate the integral using contour integration if one could find the right function and contour combination.

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Where did you find this monster? – nbubis Feb 21 '13 at 14:40
@nbubis: the bark is worse than the bite. It oddly resembles the other integral he posed which I just addressed. Look at the result: it is pretty simple, all things considered. I would use the same contour that I did for the previous integral. See this math.stackexchange.com/questions/309954/… – Ron Gordon Feb 21 '13 at 14:46

This one is similar to this one, and uses the same contour. That is, consider

$$\oint_C dz \frac{e^{r \arctan{a z}} + e^{-r \arctan{a z}}}{1+z^2} \exp{\left [i \frac{r}{2} \log{(1+a^2 z^2)} \right ]}$$

where $a>0$ and $C$ is a contour that is a semicircle in the upper half plane, except that it detours up just to the left of the imaginary axis to $z=i/a$, around that point, and the back down just to the right of the imaginary axis to the real axis, where it continues along the semicircle. This detour is needed to avoid the branch point at $z=i/a$.

In this way, note that there is a pole within $C$ at $z=i$ only when $a>1$. When $a<1$, then there is no pole within $C$ and the integral is zero. So for $a>1$ we have the integral along the real axis is equal to $i 2 \pi$ times the residue at the pole $z=i$:

$$\int_{-\infty}^{\infty} dx \frac{e^{r \arctan{a x}} + e^{-r \arctan{a x}}}{1+x^2} \exp{\left [i \frac{r}{2} \log{(1+a^2 x^2)} \right ]} = i 2 \pi \frac{e^{i r \mathrm{arctanh}{a}}+e^{-i r \mathrm{arctanh}{a}}}{2 i} \exp{\left [i \frac{r}{2} \log{(1-a^2)} \right ]}$$

Again, use the fact that

$$\mathrm{arctanh}(a) = \frac{1}{2} \log{\left ( \frac{1+a}{1-a} \right )}$$

and we get

$$\int_{-\infty}^{\infty} dx \frac{e^{r \arctan{a x}} + e^{-r \arctan{a x}}}{1+x^2} \exp{\left [i \frac{r}{2} \log{(1+a^2 x^2)} \right ]} = 2 \pi \cos{r \log{(1+a)}}$$

We finish this by considering the same integral, complex conjugated, using a contour in the lower half-plane; the results are identical. Therefore,

$$\int_{-\infty}^{\infty} dx \frac{e^{r \arctan{a x}} + e^{-r \arctan{a x}}}{1+x^2} \cos{\left [ \frac{r}{2} \log{(1+a^2 x^2)} \right ]} = 2 \pi \cos{r \log{(1+a)}}$$

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There is a curious formula attributed to Abel that states if $f(a+z)$ can be expanded in a series of the form $f(a+z) = \sum_{n=0}^{\infty} c_{n} e^{-nz}$ whether $z$ be real or imaginary, then for $b >0$, $$\int_{0}^{\infty} \frac{f(a+ibt)+f(a-ibt)}{1+t^{2}} \, dt = \pi f(a+b).$$

If we assume $\sum_{n=0}^{\infty} c_{n}$ converges absolutely, the proof is fairly straightforward.

\begin{align} \int_{0}^{\infty} \frac{f(a+ibt)+f(a-ibt)}{1+t^{2}} \ dt &= 2 \int_{0}^{\infty}\frac{1}{1+t^{2}} \sum_{n=0}^{\infty} c_{n} \cos(nbt) \, dt \\ &= 2 \sum_{n=0}^{\infty} c_{n} \int_{0}^{\infty} \frac{\cos (nbt)}{1+t^{2}} \, dt \\ &= \pi \sum_{n=0}^{\infty} c_{n} e^{-nb} \\ &= \pi f(a+b) . \end{align}

It's not particularly clear, however, what sort of functions we're talking about here.

But if we assume that $\cos \Big(r \log (1+z)\Big)$ is such a function, we get

\begin{align} &\int_{0}^{\infty} \frac{\cos \Big(r \log(1+iat) \Big) + \cos \Big(r \log(1-iat) \Big)}{1+t^{2}} \ dt \\ &= \int_{0}^{\infty} \frac{\cos \Big( \frac{r}{2} \log(1+a^{2}t^{2})+ir \arctan at \Big) + \cos \Big( \frac{r}{2} \log(1+a^{2}t^{2})-ir \arctan at \Big)}{1+t^{2}} \, dt \\ &= 2 \int_{0}^{\infty} \frac{\cosh (r \arctan at)}{1+t^{2}} \, \cos \Big(\frac{r}{2} \log(1+a^{2}t^{2})\Big) \, dt \\ &= \pi \cos \Big(r \log(1+a) \Big).\end{align}

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Note that \begin{align} &\bracks{\exp\pars{r\arctan\pars{ax}} + \exp\pars{-r \arctan(ax)}}\, \cos\pars{{r \over 2}\,\ln\pars{1 + a^{2}x^{2}}} \\[3mm]&=2\cosh\pars{r\arctan\pars{ax}} \cosh\pars{\ic\,{r \over 2}\,\ln\pars{1 + a^{2}x^{2}}} \\[3mm]&=\cosh\pars{r\arctan\pars{ax} +\ic\,{r \over 2}\,\ln\pars{1 + a^{2}x^{2}}} \\[3mm]&+\cosh\pars{r\arctan\pars{ax} -\ic\,{r \over 2}\,\ln\pars{1 + a^{2}x^{2}}} \\[3mm]&=2\Re \cosh\pars{r\arctan\pars{ax} +\ic\,{r \over 2}\,\ln\pars{1 + a^{2}x^{2}}} \\[3mm]&=2\Re \cosh\pars{r\,{\ic \over 2}\,\ln\pars{1 - \ic ax \over 1 + \ic ax} +\ic\,{r \over 2}\,\bracks{\ln\pars{1 - \ic ax} + \ln\pars{1 + \ic ax}}} \\[3mm]&=2\Re\cosh\pars{\ic r\ln\pars{1 - \ic ax}} \end{align}

such that \begin{align} &\!\!\!\!\!\!\!\color{#66f}{\large\int_{-\infty}^{\infty}\!\!{% \exp\pars{r\arctan\pars{ax}} + \exp\pars{-r \arctan(ax)} \over 1+x^2}\, \cos\pars{\!{r \over 2}\,\ln\pars{1 + a^{2}x^{2}}\!}\,\dd x} \\[3mm]&=2\Re\int_{-\infty}^{\infty} {\cosh\pars{\ic r\ln\pars{1 - \ic\verts{a}x}} \over 1 + x^{2}}\,\dd x =2\Re\braces{2\pi\ic\, {\cosh\pars{\ic r\ln\pars{1 - \ic\verts{a}\bracks{\ic}}} \over \ic + \ic}} \\[3mm]&=2\pi\,\Re\bracks{\cosh\pars{\ic r\ln\pars{1 + \verts{a}}}} =\color{#66f}{\large 2\pi\cos\pars{r\ln\pars{1 + \verts{a}}}} \end{align}

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