I've been working on a maths problem as part of my Physics PhD; but have been stumped by the following integral. All I need to know is a numeric approximation to the integral (along with an estimation of the error); but if an analytic solution exists, that would be fantastic.
$$\int_{-\infty}^{\infty} G(x,\sigma) e^{(-i(\delta x – \nu \exp(-\kappa x)/\kappa))}dx$$
where $G(x,\sigma)$ is a Gaussian distribution in $x$ with standard deviation $\sigma$ and mean zero; and where $\delta$, $\nu$ and $\kappa$ are real constants.
I have tried several approaches, including method of steepest descent/stationary phase and assymptotic expansions in $x$; but every method apart from a brute force ODE solver or polygon approach [both of which are extremely slow, such that I have not actually allowed them to complete] have proven to be very unreliable. I have considered contour integration also, but since this integral does not have any poles, I am not sure how helpful this would be.
I have not found there to be any simplifying relationships between the real constants, but here are some typical values:
$$
\begin{eqnarray}
\delta&\approx&5\times10^6\\
\nu&\approx&5\times10^6\textrm{ And slightly different from $\delta$}\\
\kappa&\epsilon& \mathbb{R^+}\\
\sigma&\gg&\frac{1}{\kappa}
\end{eqnarray}
$$
Anyway, if anyone has any ideas on how such an integral might be solved, I would be ever so grateful.
Best Answer
The reason you can't use the stationary phase technique here is that there's no point of stationary phase – the derivative of the phase
$$ \begin{align} \phi(x)&=\delta x-\nu\exp(-\kappa x)/\kappa\;, \\ \phi'(x)&=\delta+\nu\exp(-\kappa x)\;, \end{align} $$
is everywhere positive. You can turn that to your advantage though, since it allows you to change variables to $\phi$, yielding
$$ I=\int_{-\infty}^\infty \frac{G(x(\phi),\sigma)}{\phi'}\exp(-\mathrm i\phi)\,\mathrm d\phi\;. $$
This is a Fourier component of a slowly varying function. For positive $x$ we have $\phi\sim\delta x$, whereas for negative $x$ we have $\phi\sim-\nu\exp(-\kappa x)/\kappa$ and thus $x/\sigma\sim\log(-\phi)/(\kappa\sigma)$, so the integral should decay exponentially with $\delta$ and $\kappa\sigma$. You can get upper bounds for its magnitude by integrating by parts; the boundary terms vanish because of the Gaussian, so you get
$$ |I|=\left|\int_{-\infty}^\infty \frac{G(x(\phi),\sigma)}{\phi'}\exp(-\mathrm i\phi)\,\mathrm d\phi\right|\le \int_{-\infty}^\infty\left|\frac{\mathrm d^n}{\mathrm d\phi^n}\frac{G(x(\phi),\sigma)}{\phi'}\right|\,\mathrm d\phi $$
for all $n\in\mathbb N$. Since each derivative yields a large inverse factor, this should lead to very small bounds on the magnitude of the integral.
These bounds are easier to evaluate if we transform back to $x$. For instance, for $n=0$ we have
$$ |I|\le\int_{-\infty}^\infty G(x,\sigma)\,\mathrm dx=1\;, $$
for $n=1$, using $|\phi'|\lt\delta$ and $|\phi''/\phi'|\lt\kappa$,
$$ \begin{align} |I|&\le\int_{-\infty}^\infty\left|\frac{G'(x(\phi),\sigma)}{\phi'^2}-\frac{G(x(\phi),\sigma)\phi''}{\phi'^3}\right|\,\mathrm d\phi \\ &=\int_{-\infty}^\infty\left|\frac{G'(x,\sigma)}{\phi'}-\frac{G(x,\sigma)\phi''}{\phi'^2}\right|\,\mathrm dx \\ &\le\frac1\delta\int_{-\infty}^\infty\left(|G'(x,\sigma)|+\kappa|G(x,\sigma)|\right)\,\mathrm dx \\ &=\frac1\delta\left(2G(0,\sigma)+\kappa\right) \\ &=\frac1\delta\left(\frac2{\sqrt{2\pi}\sigma}+\kappa\right)\;, \end{align} $$
and for $n=2$, using $|\phi'''/\phi'|\le\kappa^2$,
$$ \begin{align} |I|&\le\int_{-\infty}^\infty\left|\frac{G''}{\phi'^3}-3\frac{G'\phi''}{\phi'^4}+3\frac{G\phi''^2}{\phi'^5}-\frac{G\phi'''}{\phi'^4}\right|\,\mathrm d\phi \\ &=\int_{-\infty}^\infty\left|\frac{G''}{\phi'^2}-3\frac{G'\phi''}{\phi'^3}+3\frac{G\phi''^2}{\phi'^4}-\frac{G\phi'''}{\phi'^3}\right|\,\mathrm dx \\ &\le\frac1{\delta^2}\int_{-\infty}^\infty\left(|G''|+3\kappa |G'|+3\kappa^2|G|+\kappa^2|G|\right)\,\mathrm dx \\ &\le\frac1{\delta^2}\left(\frac2{\sigma^2}+\frac{6\kappa}{\sqrt{2\pi}\sigma}+4\kappa^2\right)\;. \end{align} $$
We can go on like this until the cows come home, that is, until the numerical factors from the derivatives begin to overwhelm the factors of $\kappa/\delta$, so if $\kappa\ll\delta$, the integral is for all intents and purposes zero. If $\kappa\sim\delta$, you may have to work a bit harder to get better bounds.