In Marsden / Hoffman, an exercise (14 on p446) asks for an asymptotic analysis of the Gaussian integral $$\int_{\gamma} \textrm{e}^{-\zeta^{2}}d\zeta$$ along the contour $\gamma$ defined by the vertical line from $x$ to $x + ix$ with $x \in \mathbb{R}$.
By parameterising the contour $\gamma : [0, x] \rightarrow \mathbb{C}$ by $\gamma(t) = x + it$ I came to $$I(x) := \textrm{e}^{-x^{2}}\int_{0}^{x}\textrm{e}^{-2ixt + t^{2}}dt$$ and I would like to show this vanishes as $x \rightarrow \infty$ along the real axis.
I found a way tha t works by considering the absolute value of the integral and using the trick (the same trick works even without the absolute value) $$ \textrm{e}^{t^{2}} = \frac{1}{2t}\frac{d}{dt} \textrm{e}^{t^{2}}$$ which allows an integration by parts that gives $$|I(x)| \leqslant \frac{1}{2x} + \textrm{e}^{-x^{2}}\int_{1}^{x} dt \frac{\textrm{e}^{t^{2}}}{2t^{2}} + \mathcal{O}(\textrm{e}^{-x^{2}})$$ where the trailing factor comes from the region of integration $t \in [0, 1]$.
To write $|I(x)| \sim \frac{1}{2x}$ but for this to count as an asymptotic relation the remaining integral must be $\textit{o}(\frac{1}{x^{2}})$. This seems as difficult to show from the integral as it was to show the leading behaviour of $|I(x)|$ in the first place. I found two methods:
- Integrate by parts a second time to get $\textrm{e}^{-x^{2}}\int_{1}^{x} dt \frac{\textrm{e}^{t^{2}}}{2t^{2}} = \frac{1}{4x^{3}} + \frac{3}{4}\textrm{e}^{-x^{2}} \int_{0}^{x}\frac{\textrm{e}^{t^{2}}} {t^{4}}dt$ and I can show the final integral is less than $\frac{3}{4x^{3}}$.
- Use L'Hopital on $\lim_{x\rightarrow \infty} \frac{\int_{0}^{x}\frac{\textrm{e}^{t^{2}}} {t^{2}}dt}{\textrm{e}^{x^{2}} / x}$ to get limit zero.
Question: Is there a better or easier way to bound this integral from its definition?
I would prefer, however, to apply some other methods from the relevant chapter such as steepest descent or stationary phase. Yet there are various problems I'd like some help understanding:
- Steepest descent using change of variables $t = yx$ for $$ x\textrm{e}^{-x^{2}}\int_{0}^{1} \textrm{e}^{x^{2}(t^{2} -2it)}$$ whose exponent has a stationary point at $t = i$.
Question: Am I free to deform the contour in anyway required to pass through the stationary point in such a way that $\mathcal{Im}(x^{2} (\zeta^{2} - 2i\zeta))$ is constant around $\zeta = i$?
Problem: If I do this, I get that the asymptotic behaviour of $I(x) \sim 1$. What has gone wrong here?
- Stationary phase with $\textrm{e}^{-x^{2}}\int_{0}^{x} \textrm{e}^{-2itx}\textrm{e}^{t^{2}}$. Now the problem is that the stationary point is at $t = 0$. In any case, by my calculation stationary phase gives the (incorrect) asymptotic behaviour $I(x) \sim \textrm{e}^{x^{2}}/\sqrt{x}$ (actually since the phase exponent is linear in $t$ the second derivative is vanishing so perhaps this result is incorrectly obtained).
Can anyone confirm these approaches and point out where things have gone wrong?
The problem with OP's approach is essentially that there isn't a stationary point nearby. Instead let us employ the same strategy as in my Math.SE answer here. Write OP's integral as
$$\begin{align}I(x) ~=~&\int_{x}^{x+ix} \! \mathrm{d}z~e^{-z^2}~=~[e^{-a^2}J(a)]^{a=x}_{a=x+ix}\cr ~\stackrel{(D)}{=}~&\left(\frac{e^{-x^2}}{2x} -\frac{e^{-2ix^2}}{2x(1+i)}\right)\left( 1+O(x^{-2}) \right),\end{align}\tag{A}$$
where
$$J(a)~:=~\int_{\mathbb{R_+}} \!\mathrm{d}z~e^{-z(z+2a)}.\tag{B}$$
Next change integration variable
$$u~=~z(z+2a)\qquad \Leftrightarrow\qquad z~=~\sqrt{u+a^2}-a.\tag{C}$$
Then
$$ \begin{align}J(a) ~=~&\int_{\mathbb{R_+}} \!\mathrm{d}u~ \frac{e^{-u}}{2\sqrt{u+a^2}} ~=~\frac{1}{2a}\int_{\mathbb{R_+}} \!\mathrm{d}u~ \frac{e^{-u}}{\sqrt{1+\frac{u}{a^2}}}\cr ~=~&\frac{1}{2a}\int_{\mathbb{R_+}} \!\mathrm{d}u\left( 1- \frac{u}{2a^2} +\frac{3u^2}{8a^4}+O(a^{-6}) \right)\cr ~=~&\frac{1}{2a}\left( 1- \frac{1}{2a^2} +\frac{3}{4a^4}+O(a^{-6}) \right).\end{align} \tag{D}$$