# 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

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

Can anybody tell me how to deal with such integrals?

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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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Here is an approach that doesn't use contour integration.

Assuming that $f(a+z)$ can be expanded in the form $$f(a+z) = C_{0}+C_{1}e^{-z}+C_{2}e^{-2z} + \ldots$$

then for $b > 0$

$$\int_{0}^{\infty} \frac{f(a+ibt)-f(a-ibt)}{1+t^{2}} \ dt$$

$$= \int_{0}^{\infty}\frac{\Big(C_{0} + C_{1} (\cos bt - i \sin bt) + \ldots \Big) + \Big(C_{0} + C_{1} (\cos bt - i \sin bt) + \ldots \Big)}{1+t^{2}} \ dt$$

$$= 2 \int_{0}^{\infty}\frac{C_{0} + C_{1} \cos bt +C_{2} \cos 2bt + \ldots}{1+t^{2}} \ dt$$

$$= 2 \Big( C_{0} \frac{\pi}{2} + C_{1} \frac{\pi}{2} e^{-b} + C_{2} \frac{\pi}{2} e^{-2b} + \ldots \Big) = \pi \Big(C_{0} + C_{1}e^{-b} +C_{2}e^{-2b} + \ldots \Big)$$

$$= \pi f(a+b)$$

Now let $f(z) = \cos (r \log z)$.

Then

$$\int_{0}^{\infty} \frac{f(1+iat)+f(1-iat)}{1+t^{2}} \ dt = \int_{0}^{\infty} \frac{\cos \Big(r \log(1+iat) \Big) + \cos \Big(r \log(1-iat) \Big)}{1+t^{2}} \ dt$$

$$=\pi \cos \Big(r \log(1+a) \Big)$$

But

$$\cos \Big(r \log(1+iat) \Big) + \cos \Big(r \log(1-iat) \Big)$$

$$= \cos \Big( \frac{r}{2} \log(1+a^{2}t^{2})+ir \arctan at \Big) + \cos \Big( \frac{r}{2} \log(1+a^{2}t^{2})-ir \arctan at \Big)$$

$$= 2 \cos \Big(\frac{r}{2} \log(1+a^{2}t^{2})\Big) \cosh(r \arctan at)$$

Therefore,

$$\int_{0}^{\infty} \frac{\cosh (r \arctan at)}{1+t^{2}} \cos \Big( \frac{r}{2} \log(1+a^{2}t^{2}) \Big) \ dt = \frac{\pi}{2} \cos \Big( r\log(1+a) \Big)$$

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