Finding the asymptotic behaviour of an integral

Question

I want to find the full asymptotic behaviour of the integral

$$I = \int_1^\infty (t^2-1)^{-1/2}e^{-xt}\,{\rm d}t$$

at $x\to\infty$. So firstly we note that the integrand doesn't exist at $t=1$ so we must attempt to remove the "singularity". Letting $t=\cosh{(u)}$ gives

$$I = \int_0^\infty e^{-x\cosh{u}}\,{\rm d}u$$

then recall that

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\qquad \cosh{x}=\sum_{m=0}^\infty \frac{x^{2m}}{(2m)!}$$

which both converge for all $x\in\mathbb{R}$. Using these we find that

$$I=\int_0^\infty \sum_{n=0}^\infty\frac{(-x)^n}{n!}\left(\sum_{m=0}^\infty \frac{u^{2m}}{(2m)!}\right)^n\,{\rm d}x$$

Now how would I progress?

Answer 1

Sub $t=\cosh{u}$; then

$$I(x) = \int_0^{\infty} du \, e^{-x \cosh{u}} $$

Apply Laplace's method. Note that, while

$$\cosh{u} = 1 + \frac{u^2}{2!} + \sum_{k=2}^{\infty} \frac{u^{2 k}}{(2 k)!} $$

we Taylor expand the exponential only for the higher order terms in the sum in the limit as $x \to \infty$. Thus, for example, to order $2 N$:

$$I(x) = e^{-x} \int_0^{\infty} du \, e^{-x u^2/2} \left [1+ a_4 u^4 + a_6 u^6 + \cdots + O(u^{2 N}) \right ] $$

Now, the coefficients $a_4$, $a_6$, and so on would need to be computed via Taylor expansion of the exponential of the above sum. Use the fact that

$$\int_0^{\infty} du \, u^{2 k} e^{-x u^2/2} = \frac12 \left (\frac{2}{x} \right )^{k+1/2} \Gamma \left ( k+\frac12 \right ) $$

and you can derive the full asymptotic expansion to as many terms as you need.

Answer 2

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$\ds{I \equiv \int_{1}^{\infty}\pars{t^{2} - 1}^{-1/2}\expo{-xt}\,\dd t:\ ?}$.

I'll 'remove' the integrable singularity at $\ds{t = 1}$. Namely, $\ds{t \equiv 1 + \xi^{2}}$: \begin{align} I & \equiv \int_{1}^{\infty}\pars{t^{2} - 1}^{-1/2}\expo{-xt}\,\dd t = 2\expo{-x}\int_{0}^{\infty}{\expo{-x\xi^{2}} \over \root{\xi^{2} + 2}}\,\dd\xi \\[5mm] &= \root{2}\expo{-x}\int_{0}^{\infty}\expo{-x\xi^{2}} \exp\pars{-\,{1 \over 2}\ln\pars{1 + {1 \over 2}\,\xi^{2}}}\,\dd\xi \end{align}

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When $\ds{x \to \infty}$, the 'main contribution' to the integral comes from $\ds{\xi \gtrsim 0}$ such that ( by means of the Laplace Method ): \begin{align} I & = \root{2}\expo{-x}\int_{0}^{\infty}\expo{-x\xi^{2}} \bracks{1 - {1 \over 4}\,\xi^{2} + {3 \over 32}\,\xi^{4} - {5 \over 228}\,\xi^{6} + \mrm{O}\pars{\xi^{8}}}\,\dd\xi \end{align}

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For instance, up to $\ds{\large\color{#f00}{3}}$ terms $\ds{\pars{~\mbox{note that}\ \int_{0}^{\infty}\expo{-x\xi^{2}}\xi^{\mu}\,\dd\xi = {\Gamma\pars{\bracks{\mu + 1}/2} \over 2x^{\pars{\mu + 1}/2}}~}}$: \begin{align} I & \equiv \int_{1}^{\infty}\pars{t^{2} - 1}^{-1/2}\expo{-xt}\,\dd t \sim {\root{2} \over 2}\expo{-x}\bracks{{\Gamma\pars{1/2} \over x^{1/2}} - {1 \over 4}\,{\Gamma\pars{3/2} \over x^{3/2}} + {3 \over 32}\,{\Gamma\pars{5/2} \over x^{5/2}}}\ \mbox{as}\ x \to \infty \\[5mm] I & \equiv \int_{1}^{\infty}\pars{t^{2} - 1}^{-1/2}\expo{-xt}\,\dd t\,\,\, \bbx{\ds{\sim \root{\pi \over 2}\expo{-x}\pars{{1 \over x^{1/2}} - {1 \over 8x^{3/2}} + {9 \over 128x^{5/2}}}\ \mbox{as}\ x \to \infty}} \end{align}