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
2

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

$$
int_0^1 e^-x(1-u^2),du = frac12 int_0^1 e^-xv (1-v)^-1/2,dv.
$$

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 $v=infty$:

$$
beginalign
int_0^1 e^-x(1-u^2),du &= frac12 int_0^1 e^-xv (1-v)^-1/2,dv \
&approx frac12 sum_k=0^infty binom-1/2k (-1)^k int_0^infty e^-xv v^k,dv \
&= frac12 sum_k=0^infty binom-1/2k frac(-1)^k k!x^k+1
endalign
$$

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

$$
I(v) = int_0^a e^-xvg(v),dv approx sum_k=0^infty fracg^(k)(0)x^k+1
$$

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

$$
g(v) = sum_k=0^N fracg^(k)(0)k! v^k + fracg^(N+1)(v^*)(N+1)! v^N+1. tag1
$$

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

$$
I(v) = int_0^delta e^-xvg(v),dv + int_delta^a e^-xvg(v),dv = I_1(v) + I_2(v).
$$

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

$$
I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + frac1(N+1)! int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv.
$$

The last integral can be bounded by

$$
leftlvert int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv rightrvert leq left( sup_0 < v < delta leftlvert g^(N+1)(v) rightrvert right) int_0^infty e^-xv v^N+1,dv = fractextconst.x^N+2.
$$

Thus

$$
I_1(v) = I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + O!left(x^-N-2right)
$$

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

$$
beginalign
int_0^delta e^-xv v^k ,dv &= int_0^infty e^-xv v^k,dv - int_delta^infty e^-xv v^k,dv \
&= frack!x^k+1 + O!left(e^-delta xright),
endalign
$$

and substitute these into the sum to get

$$
I_1(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$

and hence

$$
I(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$

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

$$
I(v) approx sum_k=0^infty fracg^(k)(0)x^k+1
$$

as $x to infty$.

The integral can be solved in closed form in terms of the Dawson function.

$$f(x)=F(sqrtx)/sqrtx sim frac12x+frac14x^2+frac38x^3 +...$$
Mathematica was used to perform the asymptotic expansion for $s to infty.$

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