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My goal is to expand the function $$\begin{align*} f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y. \end{align*}$$ from this question asked by me: Taylor expansion of a gaussian integral.

However, this time I want to expand for large $x\gg1$. I should get an answer of the following form: \begin{align*} f_N(x) = \sum^N_{n = 0}a_n x^{\frac{-(2n+1)}{4}}. \end{align*} I have tried expanding the exponential in the integrand using the classical taylor expansion of $e$, but this lead me to nowhere. I have no clue how to begin this expansion so any hint to get me started would be very helpful :).

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To get an expansion for $x>>1$, let's first make substitution $y = x^{-\frac14}t$ in the integral: \begin{align} f(x) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^4}{24} - \frac{t^2}{2\sqrt{x}}} \frac{dt}{\sqrt[4]{x}} = \frac{x^{-\frac14}}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{t^4}{24}} \sum_{n=0}^\infty \frac{1}{n!}\left(- \frac{t^2}{2\sqrt{x}}\right)^n dt \\ &= \sum_{n=0}^\infty \frac{(-1)^nx^{-\frac14-\frac{n}{2}}}{\sqrt{2\pi} 2^n n!} \int_{-\infty}^\infty t^{2n}e^{-\frac{t^4}{24}} dt \end{align} For these integrals we have, with substitution $t = \sqrt[4]{24u} $ \begin{align} \int_{-\infty}^\infty t^{2n}e^{-\frac{t^4}{24}} dt &= 2 \int_0^\infty t^{2n}e^{-\frac{t^4}{24}} dt = \\ &= 2 (24)^{\frac{2n+1}{4}}\int_0^\infty u^{\frac{2n-3}{4}}e^{-u} du = \\ &= 2 (24)^{\frac{2n+1}{4}} \Gamma(\frac{2n+1}{4}) \end{align} where $\Gamma$ is the Euler's gamma function. In total, we get $$ f(x) = \sum_{n=0}^\infty \frac{(-1)^n \Gamma(\frac{n}{2}+\frac14)}{\sqrt{2\pi} 2^{n-1} n!} (\frac{x}{24})^{-\frac{2n+1}{4}} $$ It can be shown that this series is convergent for all $x>0$.

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  • $\begingroup$ Thank you very much ! One question though: why is it justified to switch the integral and the infinite sum ? $\endgroup$ Mar 1 '21 at 17:22
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    $\begingroup$ It can be proven by checking the finite expansion $$ e^{-\frac{t^2}{2\sqrt{x}}}= \sum_{n=0}^{N-1} \frac{1}{n!}\left(- \frac{t^2}{2\sqrt{x}}\right)^n + R_N(t,x) $$ with the remainder $$ R_N(t,x) = \frac{1}{N!}\left(- \frac{t^2}{2\sqrt{x}}\right)^N e^{-\frac{t^2}{2\sqrt{x}}\theta_N(t,x)} $$ for some $\theta_N(t,x) \in [0,1]$. Using the bound $$ |R_N(t,x)| \le \frac{1}{N!}\left(\frac{t^2}{2\sqrt{x}}\right)^N $$ it can be proven that the integral with the remainder goes to $0$ as $N\to\infty$. $\endgroup$ Mar 1 '21 at 17:56

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