I am interested in calculating the following integral over an interval $[0,T]$: $$I(c)\triangleq\int_0^T e^{-ax}\Phi\left(b\sqrt{x}+\frac{c}{\sqrt{x}}\right)\textrm{d}x$$ for $a,b,c\in\mathbb{R}$ and where $\Phi$ is the cumulative distribution function of a normal variable. Because: $$0<e^{-ax}\Phi\left(b\sqrt{x}+\frac{c}{\sqrt{x}}\right)<e^{-ax}$$ we know the integral converges over $[0,T]$. However it does not seem possible to determine a closed-form solution unless $c=0$ (for which I have already determined a closed-form expression).
Instead, I am trying to find an approximation at least for the case where $c\sim0$. Because we are integrating with respect to $x$, I thought of making a Taylor expansion around $c$ then integrating this expression in order to obtain: $$\tag{1}I(c)=\int_0^Te^{-ax}\left(\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}c^k\right)\textrm{d}x$$ where I define: $$f(c)\triangleq \Phi\left(g(c)\right),\qquad g(c)\triangleq b\sqrt{x}+\frac{c}{\sqrt{x}}$$ One can easily show by recurrence or using Faà di Bruno's formula that for $n\geq1$: \begin{align} f^{(n)}(c) &=\frac{1}{x^{n/2}}\varphi^{(n-1)}\left(g(c)\right)\\ &=\frac{(-1)^{n-1}}{x^{n/2}}\varphi\left(g(c)\right)\text{He}_{n-1}\left(g(c)\right) \end{align} where $\varphi$ is the normal density function and $\text{He}_{n}$ designates the $n$th probabilistic Hermite polynomial. If we take the case $n=3$ and evaluate at $c=0$ we have: \begin{align} f^{(3)}(0) &=\frac{1}{x^{3/2}}\varphi\left(b\sqrt{x}\right)(b^2x-1)\\ &=\varphi\left(b\sqrt{x}\right)\left(\frac{b^2}{\sqrt{x}}-\frac{1}{x^{3/2}}\right)\\ \end{align} However the integral: $$J_3\triangleq\int_0^T\frac{e^{-ax}}{x^{3/2}}\varphi(b\sqrt{x})\textrm{d}x$$ diverges because for $x\in[0,T]$: $$0<\frac{M}{x^{3/2}}<\frac{e^{-ax}}{x^{3/2}}\varphi(b\sqrt{x})$$ for some $M>0$ e.g. $M=\varphi(b\sqrt{T})$ if $a\leq0$ and $M=e^{-aT}\varphi(b\sqrt{T})$ if $a>0$. In fact one can see that the following integrals diverge for any $k>2$: $$J_k=\int_0^T\frac{e^{-ax}}{x^{k/2}}\varphi\left(b\sqrt{x}\right)\text{He}_{k-1}\left(b\sqrt{x}\right)\textrm{d}x$$ This raises the following questions:
- Is the approximation $(1)$ justified? In principle, given the left-hand side of $(1)$ is convergent, the approximation on the right-hand side should be convergent and thus valid. Numerical tests with the first two terms in the expansion show this seems legit. Does $(1)$ simply imply the integrand series is convergent?
- Can we invert integration and summation in $(1)$? If the series is convergent, then the approximation integral is well-defined but we cannot integrate piecewise (at least for $k>2$); however my numerical test does show in practice this seems to work.
If my procedure does not hold well theoretically, is there any hints on alternative ways? Asymptotic methods for integrals such as Watson's Lemma or Laplace's Method would be applicable to my integral $I$ however these methods usually require $a\rightarrow\infty$ but in my case $a$ is fixed or in any case close to $0$ so they do not seem applicable. Another alternative I have thought of is to split $I$ in two domains $[0,\varepsilon]$ and $[\varepsilon,T]$; apply my previous procedure to the second domain, then use some approximation over $x$ near $0$ for the whole integrand on the first domain, integrate and show that it goes to $0$.
Any hints are appreciated.
Assume wlg $T=1;b=1.$ Consider the case $0\ll c\ll 1$
For $$0\lt x \lt c^2$$ we have $${c \over \sqrt x} \gg \sqrt x$$ and $${\sqrt x} + {c \over \sqrt x} \sim {c \over \sqrt x}$$ For $$x \gt \sqrt c$$ we have $${\sqrt x} + {c \over \sqrt x} \sim \sqrt x$$
so the integral can be split into $$[0,c^2],[c^2, \sqrt c], [\sqrt c, 1] $$ where in the middle zone $[c^2, \sqrt c]$ there is no singularity, and since the range vanishes when $c\rightarrow 0$, the contribution is of lower order, leaving us to consider only $$[0,c^2], [\sqrt c, 1] .$$