first term in the asymptotic expansion using method of steepest descent I am working with the following intgral:
$\int_{0}^{\infty}t^{n}e^{-x(t+\frac{1}{t})}dt$
as $x\rightarrow \infty$
Now, I have been trying to solve this using the method of steepest descent. After finding the saddle points I think I have found the steepest descent path to be along the unit circle. However, I am not sure how I should now use that information to now deform the contour to do the asymptotic expansion. Any ideas?
 A: As @Maxim mentioned, the Laplas method can be applied to the integral, and the steepest descend is along the real axis, near $t=1$. The asymptotics can be obtained as the asymptotics of special function, because $I(x)=2K_{-n-1}(2x)$.
We can also get the asymptotics (few several terms) directly.
Let
$$I_n(x)=\int_{0}^{\infty}t^{n}e^{-x(t+\frac{1}{t})}dt$$
We can consider the generating function instead, namely
$$J(\alpha,x)=\sum_{n=0}^\infty\frac{\alpha ^n}{n!}I_n(x)=\int_{0}^{\infty}e^{\alpha t}e^{-x(t+\frac{1}{t})}dt=\int_{0}^{\infty}e^{\alpha t}e^{-xf(t)}dt$$
where $f(t)=\frac{1}{t}+t$.
Then $\,\,I_n(x)=\frac{\partial^n }{\partial\alpha^n}J(\alpha,x)\Big|_{\alpha=0}$
$f'(t)=0$ at $t=1$; therefore near this point
$$f(t)=f(1)+f'(1)(t-1)+f''(1)\frac{(t-1)^2}{2!}+f'''(1)\frac{(t-1)^3}{3!}+...$$
$$f(t)=2+(t-1)^2-(t-1)^3+(t-1)^4-=...$$
$$J(\alpha,x)=e^{-2x}\int_{0}^{\infty}e^{\alpha t}e^{-x\big((t-1)^2-(t-1)^3+(t-1)^4-+...\big)}dt=e^{-2x}\int_{-1}^{\infty}e^{\alpha (t+1)}e^{-x\big(t^2-t^3+t^4-+...\big)}dt$$
$$=\frac{e^{-2x+\alpha}}{\sqrt x}\int_{-\sqrt x}^{\infty}e^{\frac{\alpha t}{\sqrt x}}e^{-t^2}e^{\frac{t^3}{\sqrt x}-\frac{t^4}{x}+ -...}dt=\frac{e^{-2x+\alpha}}{\sqrt x}\int_{-\infty}^{\infty}\,-\,\,\frac{e^{-2x+\alpha}}{\sqrt x}\int_{-\infty}^{-\sqrt x}$$
The second integral contains an additional exponentially small factor ($\sim e^{-x}$), so we drop it and focus on the first term. Expanding the integrand near $t=0$ into the series
$$J(\alpha,x)=\frac{e^{-2x+\alpha}}{\sqrt x}\int_{-\infty}^{\infty}\Big(1+\frac{\alpha t}{\sqrt x}+\frac{\alpha^2 t^2}{2x}+...\Big)\Big(1+\frac{t^3}{\sqrt x}+\frac{t^6}{2x}-\frac{t^4}{x}+O\big(x^{-\frac{3}{2}}\big)\Big)e^{-t^2}dt$$
Using $\int_{-\infty}^{\infty}e^{-t^2}t^{2k}dt=(-1)^k\frac{\partial^k}{\partial s^k}\int_{-\infty}^{\infty}e^{-st^2}dt\Big|_{s=1}=(-1)^k\sqrt\pi\frac{\partial^k}{\partial s^k}\frac{1}{\sqrt s}\Big|_{s=1}$, integrating term by term and grouping the terms with the same power of $x$, we get
$$J(\alpha,x)= \sqrt\pi\,\,\frac{e^{-2x}}{\sqrt x}\,e^{\alpha}\Big(1+\frac{\alpha^2+3\alpha+\frac{3}{4}}{4x}+O\big(\frac{1}{x^2}\big)\Big)$$
$$I_n(x)=\frac{\partial^n }{\partial\alpha^n}J(\alpha,x)\Big|_{\alpha=0}$$
$$I(0)= \sqrt\pi\,\,\frac{e^{-2x}}{\sqrt x}\Big(1+\frac{3}{16x}+O\big(\frac{1}{x^2}\big)\Big)$$
$$I(1)= \sqrt\pi\,\,\frac{e^{-2x}}{\sqrt x}\Big(1+\frac{15}{16x}+O\big(\frac{1}{x^2}\big)\Big)$$
$$I(2)= \sqrt\pi\,\,\frac{e^{-2x}}{\sqrt x}\Big(1+\frac{35}{16x}+O\big(\frac{1}{x^2}\big)\Big)$$
$$....$$
$$I_n(x)=\sqrt\pi\,\,\frac{e^{-2x}}{\sqrt x}\Big(1+\frac{(2n+3)(2n+1)}{16x}+O\big(\frac{1}{x^2}\big)\Big)$$
The main asymptotic term of $I_n(x)$ does not depend on $n$.
