How to approximate this integral

Question

I am trying to approximate an integral of the following form $$ F_n = \int_{-1}^1(1-u^2)^{(n-3)/2}e^{Anu}du $$

(I am trying to work through this paper....https://ieeexplore.ieee.org/document/6875199)

I am just not seeing the substitution I need to use to easily see which neighborhoods of u dominate. Do I need to solve for the maximum of whats in the integral and use that as a guide? Any tips would be appreciated.

Answer 1

Consider first $$\int_{-1}^{+1} (1-u^2)^a\,e^{b u}\,du\quad\quad \text{with}\quad a>0\quad \text{and}\quad b>0$$

I think that the simplest way is to use the binomial expansion $$(1-u^2)^a=\sum_{k=0}^\infty \frac{(-a+k-1)!}{(-a-1)!\, k!}\,u^{2k}$$ $$\int_{-1}^{+1}u^{2k}\,e^{bu}\,du=-E_{-2 k}(-b)-E_{-2 k}(b)$$ $$\int_{-1}^{+1} (1-u^2)^a\,e^{b u}\,du=\sqrt{\pi } \,\,\Gamma (a+1)\,\, \, _0\tilde{F}_1\left(;\frac{2a+3}{2};\frac{b^2}{4}\right)$$ which has the good idea to simplify $$\int_{-1}^{+1} (1-u^2)^a\,e^{b u}\,du=\sqrt{\pi } \,\,\Gamma (a+1)\,\, \,\left(\frac{2}{b}\right)^{\frac{2a+1}{2}}\,\, \,I_{\frac{2a+1}{2}}(b)$$

You can also arrive to this result using the definition of the modified Bessel function of the first kind with a simple change of variable such as $u=\cos(x)$ for example.

So, back to the notations $$F_n=\sqrt{\pi } \,\,\Gamma \left(\frac{n-1}{2}\right)\,\, \,\left(\frac{2}{A n}\right)^{\frac{n-2}{2}}\,\, \,I_{\frac{n-2}{2}}(A n)$$

Now, using (have a look here)

$$\sqrt{x}\,\pi\, e^{-x}\,I_n(x)=\sqrt{\frac{\pi}{2}}\left( 1+ \frac{1-4n^2}{8x} + O\left(\frac{1}{x^2}\right)\right)$$ $$\log(F_n)=\frac{2 A-\log (A)-1}{2} n +\frac{1}{2} \log \left(\frac{\pi }{64 A}n\right)-\frac{96 A+37}{12 n}+ O\left(\frac{1}{n^2}\right)$$

Trying for $A=\pi$ and $n=50$ gives $$\log(F_{50})=102.774$$ while $$\log\Big[\int_{-1}^{+1}\left(1-u^2\right)^{47/2}\,e^{50 \pi u} \,du\Big]=100.834$$