Can you all please solve this one?
I was originally thining of $$\int_{-\infty}^{\infty} e^{-iax}\left(\frac{x^2}{n} + 1\right)^{-\frac12(n+1)} \,dx.$$
My mentor has told me that if n is odd you can calculate this and if n is even you will get to use special functions.
If n is odd. $$\int_{-\infty}^{\infty} e^{-iax}\left(\frac{x^2}{2k+1} + 1\right)^{-(k+1)} \,dx.$$ This case is done by Shashi by employing contour integration. Thanks.
If n is even $$\int_{-\infty}^{\infty} e^{-iax}\left(\frac{x^2}{2k} + 1\right)^{-\frac12(2k+1)} \,dx.$$ This is unsolved yet. \begin{align} I=(2k)^\frac{k+1}{2}\int^\infty_{-\infty}\frac{e^{-iax}}{(x^2+2k)^\frac{k+1}{2}}\,dx \end{align}
So it is enough to do the integral \begin{align} J = \int_\mathbb{R} \frac{e^{-iax}}{(x^2+2k)^\frac{k+1}{2}}\,dx \end{align} where b > $0$. Consider the contour integral:
\begin{align} \oint_C \frac{e^{-iaz}}{(z^2+b)^\frac{k+1}{2}}\,dz \end{align}
Shashi gives me an advice: since the exponent of the denominator of the first integral I have written is not an integer in the even case. One must then be careful with the branch cut and all that. So this is a problem of contour integral having branch cuts. But I don't know how to choose a proper contour for that...
I don't know if one would call this a closed form, but it gives a finite sum with $k$ terms. Okay, let's give it a try.
We assume $a\geq 0$. Rewriting the integral first. \begin{align} I=(2k+1)^{k+1}\int^\infty_{-\infty}\frac{e^{-iax}}{(x^2+2k+1)^{k+1}}\,dx \end{align} So, it is enough to do the integral $$J=\int_\mathbb{R}\frac{e^{-iax}}{(x^2+b)^{k+1}}\,dx$$ where $b>0$. Okay, consider the contour integral: \begin{align} \oint_C \frac{e^{-iaz}}{(z^2+b)^{k+1}}\,dz \end{align} where $C=C(R)$ is the famous semi-circle contour in the lower half plane with radius $R$ large enough so that the pole is inclosed. It is clear by Jordan's Lemma that the contribution of the circular part of the contour goes to zero as $R\to\infty$. Hence we have: \begin{align} \lim_{R\to\infty}\oint_{C(R)} \frac{e^{-iaz}}{(z^2+b)^{k+1}}\,dz =J \end{align} By the Residue Theorem one has: \begin{align} \lim_{R\to\infty}\oint_{C(R)} \frac{e^{-iaz}}{(z^2+b)^{k+1}}\,dz = -2\pi i \text{Res}_{z=-i\sqrt[]{b}} \frac{e^{-iaz}}{(z^2+b)^{k+1}} \end{align} Before doing the residue, define: \begin{align} f(z):=e^{-iaz},\ \ \ \ \ g(z):=(z-i\sqrt[]{b})^{-k-1} \end{align} Now the residue is: \begin{align} \text{Res}_{z=-i\sqrt[]{b}} \frac{e^{-iaz}}{(z^2+b)^{k+1}}&= \lim_{z\to -i\sqrt[]{b}} \frac{1}{k!}\left[ \frac{d^k}{dz^k}\frac{e^{-iaz}}{(z-i\sqrt[]{b})^{k+1}}\right]\\ &=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}f^{(k-j)}(-i\sqrt[]{b})g^{(j)}(-i\sqrt[]{b}) \end{align} Where we have used the Product Rule. Let us find the values of the derivatives of $f$: \begin{align} f^{(k-j)}(-i\sqrt[]{b})=(-ia)^{k-j}\exp(-a\sqrt[]{b}) \end{align} and the ones of $g$: \begin{align} g^{(j)}(-i\sqrt[]{b}) = (-1)^j\frac{(k+j)!}{k!(-2i\sqrt[]{b})^{k+j+1}} \end{align} So: \begin{align} \text{Res}_{z=-i\sqrt[]{b}} \frac{e^{-iaz}}{(z^2+b)^{k+1}}&=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}(-ia)^{k-j}\exp(-a\sqrt[]{b})(-1)^j\frac{(k+j)!}{k!(-2i\sqrt[]{b})^{k+j+1}}\\ &=ie^{-a\sqrt[]{b}}\sum_{j=0}^k\binom{k}{j}a^{k-j}\frac{(k+j)!}{k!^2(2\sqrt[]{b})^{k+j+1}} \end{align} Substituting $b=2k+1$ we finally get after simplifying things:
Surely one can do something similar for $a<0$..