Find the leading order asymptotic approximation of this integral

Question

$$I = \int_{-1}^{1}e^{xt^3}dt$$ I'm unsure how to start this problem. The question asks for leading order approximation as $x \to \infty$ and to use integration by parts. Should I substitute a sum into the integral? Use gamma functions maybe? I'm not sure. I've found that $$e^{xt^3}=\sum_{n=0}^{\infty}\frac{(xt^3)^n}{n!}$$ But I'm not sure what to do from here. Any help is appreciated!

Answer 1

$e^{xt^3}$ is increasing and positive on $t\in(-1,1)$, hence the asymptotic behaviour for large $x$ is given by

$$ O(1)+\int_{0}^{1}e^{xt^3}\,dt = O(1)+\frac{1}{3}\int_{0}^{1}t^{-2/3} e^{xt}\,dt = O(1)+\frac{e^x}{3x} \int_{0}^{1} (1-t)^{-2/3} x e^{-tx}\,dx $$ Now $(1-t)^{-2/3}$ is a positive $L^1(0,1)$ function, continuous and bounded in a right neighbourhood of the origin. $x e^{-tx}$ is a positive $L^1(\mathbb{R}^+)$ function with unit integral, converging to $\delta(x)$ in distribution. It follow that for large $x$ the given integral behaves like $$ \frac{e^{x}}{3x}\lim_{t\to 0}(1-t)^{-2/3} = \color{red}{\frac{e^x}{3x}}.$$

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Below is a plot (on the range $[1,25]$) of $\int_{-1}^{1} e^{xt^3}\, dt$ and $\frac{e^{x}}{3x}$, computed on Mathematica:

Here it is a plot of $\log\int_{-1}^{1}e^{xt^3}\,dx$ (blue) and $\log\left(\frac{e^x}{3x}\right)$ (purple) on the same interval:

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