# The asymptotic expansion of an integral of an exponential function

What is the asymptotic expansion of $f(x) := \int_0^1 e^{-x(1-u^2)}du$?

The integrand steeply declines near $u=1$. I tried to transform $u$ into something that is suitable for the method of steepest descent, but have not found an appropriate transformation. Integration by parts has not yield a satisfactory result, perhaps due to I having not found the right components to integrate.

## 2 Answers

The minimum of $1-u^2$ occurs at $u=1$, and near there we have $1-u^2 \approx 2(1-u)$, which is linear in the quantity $1-u$. This suggests we make the change of variables $1-u^2 = v$ (which is linear in $v$), giving

$$\int_0^1 e^{-x(1-u^2)}\,du = \frac{1}{2} \int_{0}^{1} e^{-xv} (1-v)^{-1/2}\,dv.$$

Then, following Watson's lemma, we can get the asymptotic expansion by expanding the subdominant term, $(1-v)^{-1/2}$, around $v=0$ and integrating term-by-term from $v=0$ to $v=\infty$:

\begin{align} \int_0^1 e^{-x(1-u^2)}\,du &= \frac{1}{2} \int_{0}^{1} e^{-xv} (1-v)^{-1/2}\,dv \\ &\approx \frac{1}{2} \sum_{k=0}^{\infty} \binom{-1/2}{k} (-1)^k \int_0^\infty e^{-xv} v^k\,dv \\ &= \frac{1}{2} \sum_{k=0}^{\infty} \binom{-1/2}{k} \frac{(-1)^k k!}{x^{k+1}} \end{align}

as $x \to \infty$, which matches the series given in the other answer.

Let's prove the special case of Watson's lemma we use in this answer. We'll assume that $g(v)$ is analytic at $v=0$ and that $\int_0^a \lvert g(v) \rvert\,dv$ exists (these are certainly true for $g(v) = (1-v)^{-1/2}$), and show that

$$I(v) = \int_0^a e^{-xv}g(v)\,dv \approx \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{x^{k+1}}$$

as $x \to \infty$. In other words, we will show that the asymptotic expansion for the integral can be obtained by expanding $g(v)$ in Taylor series around $v=0$ and integrating term-by-term.

Since $g(v)$ is analytic at $v=0$, there is a $\delta \in (0,a]$ such that $g^{(k)}(v)$ is analytic on $[0,\delta]$ for all $k$. Further, Taylor's theorem tells us that for any positive integer $N$ and any $v \in [0,\delta]$, there is a $v^* \in [0,v]$ for which

$$g(v) = \sum_{k=0}^{N} \frac{g^{(k)}(0)}{k!} v^k + \frac{g^{(N+1)}(v^*)}{(N+1)!} v^{N+1}. \tag{1}$$

We will split the integral $I(v)$ at this $\delta$, and estimate each piece separately. To this end, we define

$$I(v) = \int_0^\delta e^{-xv}g(v)\,dv + \int_\delta^a e^{-xv}g(v)\,dv = I_1(v) + I_2(v).$$

We only need a rough estimate on $I_2(v)$:

$$\lvert I_2(v) \rvert \leq \int_\delta^a e^{-xv} \lvert g(v) \rvert \,dv \leq e^{-\delta x} \int_0^a \lvert g(v) \rvert\,dv,$$

where we have assumed the last integral on the right is finite. Thus

$$I(v) = I_1(v) + O(e^{-\delta x})$$

as $x \to \infty$.

Now, from $(1)$ we have

$$I_1(v) = \sum_{k=0}^{N} \frac{g^{(k)}(0)}{k!} \int_0^\delta e^{-xv} v^k \,dv + \frac{1}{(N+1)!} \int_0^\delta e^{-xv} g^{(N+1)}(v^*) v^{N+1}\,dv.$$

The last integral can be bounded by

$$\left\lvert \int_0^\delta e^{-xv} g^{(N+1)}(v^*) v^{N+1}\,dv \right\rvert \leq \left( \sup_{0 < v < \delta} \left\lvert g^{(N+1)}(v) \right\rvert \right) \int_0^\infty e^{-xv} v^{N+1}\,dv = \frac{\text{const.}}{x^{N+2}}.$$

Thus

$$I_1(v) = I_1(v) = \sum_{k=0}^{N} \frac{g^{(k)}(0)}{k!} \int_0^\delta e^{-xv} v^k \,dv + O\!\left(x^{-N-2}\right)$$

as $x \to \infty$. Finally we reattach the tails to the integrals in the sum,

\begin{align} \int_0^\delta e^{-xv} v^k \,dv &= \int_0^\infty e^{-xv} v^k\,dv - \int_\delta^\infty e^{-xv} v^k\,dv \\ &= \frac{k!}{x^{k+1}} + O\!\left(e^{-\delta x}\right), \end{align}

and substitute these into the sum to get

$$I_1(v) = \sum_{k=0}^{N} \frac{g^{(k)}(0)}{x^{k+1}} + O\!\left(x^{-N-2}\right)$$

and hence

$$I(v) = \sum_{k=0}^{N} \frac{g^{(k)}(0)}{x^{k+1}} + O\!\left(x^{-N-2}\right)$$

as $x \to \infty$. Since $N$ was arbitrary, this is precisely the statement that $I(v)$ has the asymptotic expansion

$$I(v) \approx \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{x^{k+1}}$$

as $x \to \infty$.

• Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why. – Hans Aug 22 '18 at 8:46
• @Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $\int_0^1 e^{-xv} g(v)\,dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $\int_0^1 |g(v)|\,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong. – Antonio Vargas Aug 23 '18 at 0:17
• Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however. – Antonio Vargas Aug 23 '18 at 0:36
• @Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question. – Antonio Vargas Aug 23 '18 at 1:16
• It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer? – Hans Aug 23 '18 at 1:33

The integral can be solved in closed form in terms of the Dawson function.
$$f(x)=F(\sqrt{x})/\sqrt{x} \sim \frac{1}{2x}+\frac{1}{4x^2}+\frac{3}{8x^3} +...$$ Mathematica was used to perform the asymptotic expansion for $s \to \infty.$