Integral $\int_{-1}^{1} \frac{1}{(1+u^2)^{n/2}} \exp{\left(-2\pi \frac{a^2+b^2}{1+u^2}\right)} \exp{\left(-4\pi i ab \frac{u}{1+u^2}\right)} du $

Solution 1:

Your inclination about Bessel functions is right. Consider the integral $${{f}_{n}}\left( a,b \right)=\int\limits_{0}^{\pi /4}{{{e}^{-2\pi \left( {{a}^{2}}+{{b}^{2}} \right){{\cos }^{2}}\left( t \right)}}{{\cos }^{n-2}}\left( t \right)\cos \left( 2\pi ab\sin \left( 2t \right) \right)dt}$$ For convenience in notation I will let $n\to 2n+2$ and define two constants $\alpha =2\pi \left( {{a}^{2}}+{{b}^{2}} \right)$ and $\beta =2\pi ab$ and so
$${{f}_{n}}\left( \alpha ,\beta \right)=\int\limits_{0}^{\pi /4}{{{e}^{-\alpha {{\cos }^{2}}\left( t \right)}}{{\cos }^{2n}}\left( t \right)\cos \left( \beta \sin \left( 2t \right) \right)dt}$$ $${{f}_{n}}\left( \alpha ,\beta \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -\alpha \right)}^{k}}}{k!}}\sum\limits_{m=0}^{\infty }{\frac{{{\left( -1 \right)}^{m}}{{\beta }^{2m}}}{\left( 2m \right)!}}\int\limits_{0}^{\pi /4}{{{\cos }^{2\left( k+n \right)}}\left( t \right){{\sin }^{2m}}\left( 2t \right)dt}$$ Changing variables to $x=2t$ and re-writing $\cos \left( \tfrac{1}{2}x \right)$ yields $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{1}{{{2}^{n+1}}}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -\tfrac{1}{2}\alpha \right)}^{k}}}{k!}}\sum\limits_{m=0}^{\infty }{\frac{{{\left( -1 \right)}^{m}}{{\beta }^{2m}}}{\left( 2m \right)!}}\int\limits_{0}^{\pi /2}{{{\left( 1+\cos \left( x \right) \right)}^{k+n}}{{\sin }^{2m}}\left( x \right)dx}$$ Expand the power using binomials $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{1}{{{2}^{n+1}}}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -\tfrac{1}{2}\alpha \right)}^{k}}}{k!}}\sum\limits_{m=0}^{\infty }{\frac{{{\left( -1 \right)}^{m}}{{\beta }^{2m}}}{\left( 2m \right)!}}\sum\limits_{j=0}^{\infty }{\left( \begin{matrix} k+n \\ j \\ \end{matrix} \right)}\int\limits_{0}^{\pi /2}{\cos {{\left( x \right)}^{j}}{{\sin }^{2m}}\left( x \right)dx}$$ The remaining integral is the beta function and so $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{1}{{{2}^{n+2}}}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -\tfrac{1}{2}\alpha \right)}^{k}}}{k!}}\sum\limits_{m=0}^{\infty }{\frac{{{\left( -1 \right)}^{m}}{{\beta }^{2m}}}{\left( 2m \right)!}}\sum\limits_{j=0}^{\infty }{\left( \begin{matrix} k+n \\ j \\ \end{matrix} \right)}\frac{\Gamma \left( \tfrac{1}{2}\left( j+1 \right) \right)\Gamma \left( m+\tfrac{1}{2} \right)}{\Gamma \left( m+\tfrac{1}{2}j+1 \right)}$$ Note we have (any reasonable reference here will suffice – DLMF is nice) $$\sum\limits_{m=0}^{\infty }{\frac{{{\left( -1 \right)}^{m}}{{\beta }^{2m}}}{\left( 2m \right)!}}\frac{\Gamma \left( m+\tfrac{1}{2} \right)}{\Gamma \left( m+\tfrac{1}{2}j+1 \right)}=\sqrt{\pi }{{\left( \frac{2}{\beta } \right)}^{j/2}}{{J}_{j/2}}\left( \beta \right)$$ Therefore $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{\sqrt{\pi }}{{{2}^{n+2}}}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -\alpha /2 \right)}^{k}}}{k!}}\sum\limits_{j=0}^{\infty }{\left( \begin{matrix} k+n \\ j \\ \end{matrix} \right)}{{\left( \frac{2}{\beta } \right)}^{j/2}}\Gamma \left( \tfrac{1}{2}\left( j+1 \right) \right){{J}_{j/2}}\left( \beta \right)$$ From here we will perform the summation over k by first writing the above as $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{\sqrt{\pi }}{{{2}^{n+2}}}\sum\limits_{j=0}^{\infty }{{{\left( \frac{2}{\beta } \right)}^{j/2}}\Gamma \left( \tfrac{1}{2}\left( j+1 \right) \right){{J}_{j/2}}\left( \beta \right)}\sum\limits_{k=0}^{\infty }{\frac{\Gamma \left( n+1 \right)}{\Gamma \left( j+1 \right)\Gamma \left( n-j+1 \right)}\frac{{{\left( n+1 \right)}_{k}}}{{{\left( n-j+1 \right)}_{k}}}\frac{{{\left( -\alpha /2 \right)}^{k}}}{k!}}$$ And now we see that we have a hypergeometric function, namely $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{\sqrt{\pi }}{{{2}^{n+2}}}\sum\limits_{j=0}^{\infty }{\left( \begin{matrix} n \\ j \\ \end{matrix} \right)}{{\left( \frac{2}{\beta } \right)}^{j/2}}M\left( 1+n,1+n-j,-\tfrac{1}{2}\alpha \right)\Gamma \left( \tfrac{1}{2}\left( j+1 \right) \right){{J}_{j/2}}\left( \beta \right)$$ where $M\left( 1+n,1+n-j,-\tfrac{1}{2}\alpha \right)={}_{1}{{F}_{1}}\left( 1+n,1+n-j;-\tfrac{1}{2}\alpha \right)$ is known as Kummer’s confluent hypergeometric function of the first kind. This is not particularly nice for computation if n is an arbitrary number because the second argument in M become negative and the function doesn’t exist (well it does but only in a limit scenario, as we will soon see).
However if n is an integer then things become different. So from here on in let n be an integer and note that while M doesn’t exist when the second argument is a negative integer, in those cases we have the limit $\underset{b\to -n}{\mathop{\lim }}\,\frac{M\left( a,b,z \right)}{\Gamma \left( b \right)}=\mathbf{M}\left( a,-n,z \right)$ where M is Kummer’s confluent function of the second kind. Therefore we can break the series in two $$\begin{align} & {{f}_{n}}\left( \alpha ,\beta \right)=\frac{\sqrt{\pi }}{{{2}^{n+2}}}\sum\limits_{j=0}^{n}{\left( \begin{matrix} n \\ j \\ \end{matrix} \right)}{{\left( \frac{2}{\beta } \right)}^{j/2}}M\left( 1+n,1+n-j,-\tfrac{1}{2}\alpha \right)\Gamma \left( \tfrac{1}{2}\left( j+1 \right) \right){{J}_{j/2}}\left( \beta \right) \\ & +\frac{\sqrt{\pi }}{{{2}^{n+2}}}\sum\limits_{j=0}^{\infty }{\frac{\Gamma \left( n+1 \right)}{\Gamma \left( n+j+2 \right)}}{{\left( \frac{2}{\beta } \right)}^{\left( n+j+1 \right)/2}}\mathbf{M}\left( 1+n,-j,-\tfrac{1}{2}\alpha \right)\Gamma \left( \tfrac{1}{2}\left( n+j+2 \right) \right){{J}_{\left( n+j+1 \right)/2}}\left( \beta \right) \\ \end{align}$$ This is useful when n is large as the first finite summation yields a very good approximation and so the infinite series can be viewed as a remainder. To see this observe for large n we have $$\mathbf{M}\left( 1+n,-j,-\tfrac{1}{2}\alpha \right)\simeq {{\left( \frac{-\tfrac{1}{2}\alpha }{1+n} \right)}^{\left( 1+j \right)/2}}{{e}^{-\tfrac{1}{4}\alpha }}\frac{\Gamma \left( 2+n+j \right)}{\Gamma \left( 1+n \right)}\left( {{I}_{-j-1}}\left( 2i\sqrt{\tfrac{1}{2}\alpha \left( n+1 \right)} \right)+O\left( {{n}^{-1}} \right) \right)$$ Where ${{I}_{v}}\left( z \right)$ is the modified Bessel function of the first kind. We can ignore the negative index on the Bessel function (it comes out as positive and possibly a multiplicative term out the front) and the imaginary term in the argument turns the modified Bessel function into just the Bessel function of the first kind. Hence we will have something along the lines of $$\mathbf{M}\left( 1+n,-j,-\tfrac{1}{2}\alpha \right)\simeq {{\left( \frac{-\tfrac{1}{2}\alpha }{1+n} \right)}^{\left( 1+j \right)/2}}{{e}^{-\tfrac{1}{4}\alpha }}\frac{\Gamma \left( 2+n+j \right)}{\Gamma \left( 1+n \right)}\left( {{J}_{j+1}}\left( 2\sqrt{\tfrac{1}{2}\alpha \left( n+1 \right)} \right)+O\left( {{n}^{-1}} \right) \right)$$. Bessel functions of the first kind wiggle and decrease as their arguments get large and so we can essentially ignore it in determining the behaviour of M for large n. Note also we have $${{J}_{\left( n+j+1 \right)/2}}\left( \beta \right)\sim \frac{1}{\sqrt{\left( n+j+1 \right)\pi }}{{\left( \frac{e\beta }{n+j+1} \right)}^{\left( n+j+1 \right)/2}}$$
$$\Gamma \left( \tfrac{1}{2}\left( n+j+2 \right) \right)\simeq \sqrt{2\pi }{{e}^{-\tfrac{1}{2}\left( n+j+2 \right)}}{{\left( \tfrac{1}{2}\left( n+j+2 \right) \right)}^{\tfrac{1}{2}\left( n+j+1 \right)}}+O\left( {{n}^{-1}} \right)$$ All these asymptotic expansion lead to some rather lovely cancelations (I strongly urge you to go through it just to see how beautifully it all falls out), and we are left with something behaving like $${{f}_{n}}\left( \alpha ,\beta \right)=\frac{\sqrt{\pi }}{{{2}^{n+2}}}\sum\limits_{j=0}^{n}{\left( \begin{matrix} n \\ j \\ \end{matrix} \right)}{{\left( \frac{2}{\beta } \right)}^{j/2}}M\left( 1+n,1+n-j,-\tfrac{1}{2}\alpha \right)\Gamma \left( \tfrac{1}{2}\left( j+1 \right) \right){{J}_{j/2}}\left( \beta \right)+R\left( n \right)$$ where $$R\left( n \right)\simeq \frac{\sqrt{2\pi }{{e}^{-\tfrac{1}{4}\alpha -\tfrac{1}{2}}}}{{{2}^{n+2}}}\sum\limits_{j=0}^{\infty }{\frac{1}{\sqrt{\left( n+j+1 \right)}}{{\left( \frac{-\tfrac{1}{2}\alpha }{1+n} \right)}^{\left( 1+j \right)/2}}\left( {{J}_{j+1}}\left( 2\sqrt{\tfrac{1}{2}\alpha \left( n+1 \right)} \right)+O\left( {{n}^{-1}} \right) \right)}$$ More accurate estimates can be achieved here, but what we have now is good enough to see that for large n, and $\alpha $ such that their ratio ($\alpha \le n+1$) ensures convergence, then the remainder heads to zero (independent of $\beta $) and we can expect the finite sum to start becoming a valid approximation. In actual fact and as we can see from above, n doesn’t have to be too large (w.r.t $\alpha $) before we get good results with the finite sum. I suggest you generate some tables to see this in action, however observe the following graph below. Here we are considering ${{f}_{n}}\left( 8,3 \right)$ and so the finite sum shouldn’t start becoming good until n>8. This is exactly what we see where finite sum = blue, numerical integration = yellow. Obviously the further n goes, the better the approximation becomes.