Help with exponential integrals

Question

I'm trying to find a nice expression for the following function \begin{equation} f_k(x)=\int_0^\infty y^k (x+y) e^{-(x+y)^2} \text{d}y. \end{equation} So far I know that \begin{equation} f_k(x)=P_k(x)e^{-x^2}+Q_k(x)\text{ erfc}(x), \end{equation} where $P_k(x)$ and $Q_k(x)$ are polynomials of degree $k-1$ and $k$, respectively. However, I'm having trouble with finding general expressions for these polynomials. I'm hoping someone can help me with this.

Answer 1

By direct computation, we have: $$ P_0 = \frac{1}{2}, \qquad Q_0 = 0, $$ $$ P_1 = 0, \qquad Q_1 = \frac{\sqrt{\pi}}{4}. $$ Assuming $k>1$, integration by parts gives: $$ f_k(x) = \frac{k}{2}\int_{0}^{+\infty}y^{k-1}e^{-(x+y)^2}dy. $$ For the sake of simplicity, define: $$ g_\tau(x) = \int_{x}^{+\infty}(y-x)^\tau e^{-y^2}dy. $$ We clearly have: $$ g_\tau(x) = \int_{x}^{+\infty}(y-x)^{\tau-1}\,y e^{-y^2}dy-x\int_{x}^{+\infty}(y-x)^{\tau-1}\, e^{-y^2}dy, $$ and integrating by parts the first integral in the RHS we get: $$ g_{\tau}(x) = \frac{\tau-1}{2}g_{\tau-2}(x)-x\cdot g_{\tau-1}(x).$$ Since $f_k(x)=\frac{k}{2}g_{k-1}(x)$, $f_k(x)$ satisfies the recurrence relation: $$f_k(x) = \frac{k}{2}f_{k-2}(x) - \frac{k}{k-1}x\cdot f_{k-1}(x), $$ so $P_k(x)$ and $Q_k(x)$ can be computed in a recursive way through the same formula: $$P_k(x) = \frac{k}{2}P_{k-2}(x) - \frac{k}{k-1}x\cdot P_{k-1}(x), $$ $$Q_k(x) = \frac{k}{2}Q_{k-2}(x) - \frac{k}{k-1}x\cdot Q_{k-1}(x). $$ If we further set: $$ f_k(x) = \frac{k(-1)^k}{2^{k}}h_k(x), $$ the recursion simplifies to: $$ h_k(x)= 2x\cdot h_{k-1}(x) + 2(k-2)\cdot h_{k-2}(x). $$

Answer 2

It might help to do the change of variables $t = x+y$. Then you have $$ f_k(x) = \int_x^{\infty}(t-x)^k t e^{-t^2}dt = \sum_{l=0}^k {{k}\choose{l}} (-x)^l \int_x^{\infty}t^{k-l+1}e^{-t^2}dt $$ so you can reduce the problem to looking at finding expressions to integrals of the form $$ \int_x^{\infty} t^m e^{-t^2}dt. $$

A nice trick to try and evaluate these moments is to let $h(\lambda, x)$ and $g(\lambda,x)$ be the functions $$ h(\lambda,x) = \int_x^{\infty} e^{-\lambda t^2} dt\\ g(\lambda,x) = \int_x^{\infty} t e^{-\lambda t^2} dt $$ where you should be able then to find $h(\lambda, x)$ in terms of the error function and $g(\lambda, x)$ by a $u$-substitution. Now notice that $$ \left(-\frac{\partial}{\partial \lambda}\right)^{n}h(\lambda, x) \bigg|_{\lambda = 1} = \int_x^{\infty}t^{2n}e^{-t^2}dt \\ \left(-\frac{\partial}{\partial \lambda}\right)^{n}g(\lambda, x) \bigg|_{\lambda = 1} = \int_x^{\infty}t^{2n+1}e^{-t^2}dt $$ which will then let you find all the even and odd moments respectively by "simply" differentiating the functions $h$ and $g$ in $\lambda$ rather than doing many many integrals.

Answer 3

As suggested by Tom, you could focus on the integrals of t^m Exp[-t^2] (t from x to Infinity). The result of this integral seems to just be Gamma[(m+1)/2,x^2]/2