Integral $\int_{-\infty}^{\infty}\frac{\cos(s \arctan(ax))}{(1+x^2)(1+a^2x^2)^{s/2}}dx$

Prove that:

$$\int_{-\infty}^{\infty}{\cos\left(\vphantom{\Large A} s\ \arctan\left(\vphantom{\large A}ax\right)\right)\over \left(1 + x^{2}\right)\left(1 + a^{2}x^{2}\right)^{s/2}}\,{\rm d}x ={\pi \over \left(1 + a\right)^{s}}$$

where $a,s \in \mathbb{R}^{+}$.

This looks difficult. What would be a good start? Can we use the Residue Theorem?

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I would start with a substitution, $u = \arctan ax$ and then hit the resulting integral with residue calculus. –  mrf Feb 21 '13 at 9:57
@mrf: That's a good suggestion. I will try it. –  Integrals and Series Feb 21 '13 at 10:44
First of all, I'd have a very serious and most probably unpleasant talk with the sadist that came up with this integral. Second, perhaps the identities $$\cos\arctan x=\frac{1}{\sqrt{1+x^2}}\,\,,\,\,\sin\arctan x=\frac{x}{\sqrt{1+x^2}}$$can help a little... –  DonAntonio Feb 21 '13 at 13:53
The result should be $\displaystyle{\large{\pi \over \left(\,1 + \color{#c00000}{\left\vert\,a\,\right\vert}\,\,\right)^{\,\,\,s}}}$. –  Felix Marin Jun 28 at 1:47

I was going to add this answer a long time ago. But for some reason I didn't.

First notice that

\begin{align} \text{Re} \ (1-iax)^{-s} &= \text{Re} \ (1+a^{2}x^{2})^{-s/2} \ e^{-is\arctan (-ax)} \\ &= \text{Re} \ (1+a^{2}x^{2})^{-s/2} \ e^{i s \arctan (ax) } \\ &= \frac{\cos (s \arctan ax)}{(1+a^{2}x^{2})^{s/2}} \end{align}

Therefore,

$$\int_{-\infty}^{\infty} \frac{\cos (s \arctan ax)}{(1+x^{2})(1+a^{2}x^{2})^{s/2}} \ dx = \text{Re} \int_{-\infty}^{\infty} \frac{1}{(1+x^{2})(1-iax)^{s}} \ dx$$

Now let $$f(z) = \frac{1}{(1+z^{2})(1-iaz)^{s}}$$ and integrate around a contour that consists of the real axis and the upper half of $|z|=R$.

A simple application of the ML inequality shows that the integral vanishes along the upper half of $|z|=R$ as $R \to \infty$.

So

\begin{align} \int_{-\infty}^{\infty} \frac{\cos (s \arctan ax)}{(1+x^{2})(1+a^{2}x^{2})^{s/2}} \ dx &= \text{Re} \ 2 \pi i \ \text{Res}[f(z),i] \\ &= \text{Re} \ 2 \pi i \lim_{z \to i} \frac{1}{(z+i)(1-iaz)^{s}} \\ &= \frac{\pi}{(1+a)^{s}} \end{align}

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This may be done with a contour integration. Consider the integral

$$\oint_C dz \frac{e^{i s \arctan{a z}}}{(1+z^2)(1+a^2 z^2)^{s/2}}$$

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 consider the case when $a>1$. This contour integral is zero except along the real axis. To see this, note that along the outer semicircle of radius $R$, the integral is bounded by

$$\frac{R}{R^{2+s}} \left | \int_0^{\pi} d\theta e^{-s \mathrm{arctanh}(a R \sin{\theta})} \right | \sim \frac{1}{R^{1+s}}$$

which vanishes when $s>-1$. I will assume this for the result. The integrals along the vertical pieces cancel. The integral about the branch point $z=i/a$ may be parametrized as $z=i/a + t e^{i \phi}$, where $t \rightarrow 0$. Then we get a term for small $t$ as follows:

$$i (2 a)^{s/2} t^{1-(s/2)} \frac{e^{-s \mathrm{arctanh}(1)}}{1-1/a^2} \int_0^{2 \pi} d \phi e^{i (1-s/2) \phi} = 0$$

when $s \ne 2$. I will consider the case of $s=2$ later.

Then the integral along the real axis is equal $i 2 \pi$ times to the residue at the pole $z=i$. This is

$$i 2 \pi \frac{e^{-s \mathrm{arctanh}(a)}}{2 i (1-a^2)^{s/2}}$$

Use the fact that

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

and we finally get

$$\int_{-\infty}^{\infty} dx \frac{e^{i s \arctan{a x}}}{(1+x^2)(1+a^2 x^2)^{s/2}} = \frac{\pi}{(1+a)^s}$$

Because the result of the complex conjugate requires using the contour in the lower half-plane, we get an identical result for this. We may then say that

$$\int_{-\infty}^{\infty} dx \frac{\cos{( s \arctan{a x})}}{(1+x^2)(1+a^2 x^2)^{s/2}} = \frac{\pi}{(1+a)^s}$$

when $s>-1$ and $a>1$.

When $s=2$, we may make a substitution of $u=\arctan{a x}$ and the integral becomes

$$2 a \int_0^{\pi/2} du \frac{\cos{2 u}}{a^2 + \tan^2{u}} = \frac{\pi}{(1+a)^2}$$

so the $s=2$ case is covered.

It does seem to me that $a<1$ should also be covered by this result, but I would need to rethink the contour.

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