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

My recent research attempts got stuck in the following integral:

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

It is the most "compact" version (apart from numerical factors) I obtained from the starting problem

$$\int_{0}^{1} \frac{1}{(2x^2-2x+1)^{n/2}} e^{-A(x)\pi a^2} e^{-B(x)\pi b^2}e^{2\pi C(x) ab}\mathrm{d}x$$

where

$$A(x)=2-\frac{(2x-1)^2}{2x^2-2x+1}\qquad B(x)=\frac{1}{2x^2-2x+1}\qquad C(x)=-\frac{2x-1}{2x^2-2x+1}$$

and with $$a,b\in \mathbb{R}^n$$ ($$ab=a\cdot b$$). I tried with several substitutions (apart $$u=2x-1$$), integration by parts, complex integration, derivation with respect to a parameter ("Feynman's trick"), exploiting the complex exponential as cosine/sine, but I believe that the main obstacle to a "good" substitution is the presence of the linear term $$u$$ in the complex exponential.

I am interested in a closed form, if any.

Update 1 By symmetry arguments and with the suggested substitution I got $$\int_0^{\pi/4} (\cos \theta)^{n-2} e^{-2\pi(a^2+b^2)\cos^2 \theta} \cos(2\pi ab \sin(2\theta)) \mathrm{d}\theta$$ but I have no further ideas. I tried with several substitutions (log-tan, half-angle, etc.) but with no results. I think that some special function is involved (the last term calls for Bessel) but I am definitely not an expert. (See mathstackuser12's answer for a complete account on this).

Update 2 Before assigning the bounty I would ask if a complex integration way could be fruitful: since $$\frac{1}{1+u^2} = \frac{1}{2} \left(\frac{i}{u+i} - \frac{i}{u-i} \right)$$ and $$\overline{\frac{1}{u-i}}=\frac{1}{u+i}$$ we can write $$\int_{0}^1 \frac{1}{(1+u^2)^{n/2}}\exp{\left(-2\pi (a-b)^2 \frac{1}{1+u^2}\right)} \exp{\left(-4\pi ab \frac{i}{u+i}\right)}$$ or $$\int_{0}^1 \frac{1}{(1+u^2)^{n/2}} \exp{\left(\pi (a-b)^2 \frac{i}{u-i}\right)} \exp{\left(-(\pi(a-b)^2+4\pi ab)\frac{i}{u+i}\right)} \mathrm{d}u$$ and I wonder if there exists a useful path in the complex plane, surrounding a domain of analyticity (I tried without success), or if computing the residues in a smart way could be a good thing.‌

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. • Thank you for this very insightful contribution! I will need some time to grasp all the details. In fact, I am not interested in the asymptotic behaviour for large n, but I will think about it. Thank you again for the effort! I will wait for the bounty assignation in order to see if the latest update I added can solicitate someone's imagination - obviously the invitation is open to you too :-) – Ivan Nov 8 '17 at 14:54
• Not a problem. There is some nice mathematics wrapped up in this problem. I would remind you that the solution presented above is valid for arbitrary n,a,b (provided convergence), and not just large n - integer. It just so happens you get a 'closed-form' approximation when n is large compared with alpha. Also note that the solution is telling you a lot about your function. We now know it satisfies the confluent HG diff equation in alpha, and a form of Bessel's equation in beta. This is useful, particularly when considering contour representations. – mathstackuser12 Nov 8 '17 at 20:50
• Thanks for your interesting remarks. As said, I am not an expert of special functions (I took a first undergraduate course few years ago), I need some time to fully unwrap your answer. If you have time and you would like to add details to your points (in particular to those regarding integral representations) I will be very grateful for learning from your mastery. – Ivan Nov 9 '17 at 7:17

Hint:

$$\int_{-1}^1\dfrac{1}{(1+u^2)^\frac{n}{2}}e^{-2\pi\frac{a^2+b^2}{1+u^2}}e^{-4\pi iab\frac{u}{1+u^2}}~du$$

$$=\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\dfrac{1}{(1+\tan^2x)^\frac{n}{2}}e^{-2\pi\frac{a^2+b^2}{1+\tan^2x}}e^{-4\pi iab\frac{\tan x}{1+\tan^2x}}~d(\tan^2x)$$

$$=\int_{-\frac{\pi}{4}}^\frac{\pi}{4}e^{-2\pi(a^2+b^2)\cos^2x}e^{-2\pi iab\sin2x}\cos^{n-2}x~dx$$

• Thank you, but it seems to me that there is a mistake: the substitution you proposed is $u=\tan(x)$, so in the complex exponential you will get $-2\pi i ab \sin (2x)$. Am I missing something? – Ivan Oct 30 '17 at 13:15