2

Find the first two terms in the asymptotic expansion of

$$I(x) = \int_0^x e^{t^3} \, dt$$

as $x\rightarrow \infty$.

Attempt:

The first thing I tried was of course integration by parts:

$$\int_0^x e^{t^3} \, dt = \int_0^x \frac{1}{t^2} \cdot t^2e^{t^3} \, dt = \bigg[-\frac {1}{3t^2}e^{t^3}\bigg]_0^x - \frac 23 \int_0^x \frac{1}{t^3}e^{t^3} \, dt$$

But this doesn't work because $\frac{1}{t^2}$ blows up as $t \rightarrow 0$.

And then I realised that the original integral $I(x)$ itself blows up as $x\rightarrow \infty$, which would mean that at least the first term in the expansion should blow up as $x\rightarrow \infty$.

I took a look at other things like Watson's lemma and method of stationary phase, but they don't seem to work for things that blow up. Any hints?

Community
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glowstonetrees
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2 Answers2

6

The substitution $t=x(1-y)^{1/3}$, and Watson's lemma, give $$I(x)=\frac{x}{3}e^{x^3}\int_0^1(1-y)^{-2/3}e^{-x^3 y}\,dy\asymp\frac{e^{x^3}}{3}\sum_{n=0}^{(\infty)}\frac{\Gamma(n+2/3)}{\Gamma(2/3)x^{3n+2}}=e^{x^3}\left(\frac{1}{3x^2}+\frac{2}{9x^5}+\ldots\right).$$

metamorphy
  • 43,591
2

Split the interval into $[0,1]$ and $[1,x]$. The integral over the first interval is finite, so is swamped by the rest. Your calculations over the second integral don't blow up at 0 because 0 isn't in the interval.

Empy2
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