$ \int_c^\infty \frac{e^{-a x^2 + bx}}{x} \, dx $

Question

Suppose $a > 0$, $b$ is an arbitrary real number, and $c$ is a positive real number. Is there any closed form solution for the following integral

$$ \int_c^\infty \frac{e^{-a x^2 + bx}}{x} \, dx \text{ ?} $$

Answer 1

That is not a gaussian integral/integrand. That is: $$\int\limits_{c}^{\infty}e^{-a(x+b)^2}\,\mathrm{d}x=-\dfrac{\sqrt{{\pi}}\left(\operatorname{erf}\left(\sqrt{a}\left(c+b\right)\right)-1\right)}{2\sqrt{a}}$$ But if that's really what you meant, I don't think there is a closed form for this. If you choose $b=0$, one can write $$\int\limits_{c}^{\infty}\frac{e^{-ax^2}}{x}\,\mathrm{d}x= \dfrac{\operatorname{\Gamma}\left(0,ac^2\right)}{2}.$$ If we were alloud to set $b=d/x$ we could also look at the integral $$\int\limits_{c}^{\infty}\frac{e^{-ax^2+b}}{x}\,\mathrm{d}x= \dfrac{\operatorname{\Gamma}\left(0,ac^2\right)\mathrm{e}^d}{2}.$$

Answer 2

By changing variables, we may set one constant to unity, so it is enough to consider

$$ I=\int\limits_c^\infty \frac{dx}{x} e^{-x^2+bx} $$

Differentiating w.r.t $b$ looks promising, but leads to another tricky integral after the $x$ integration

$$ J=\int\limits_0^b dt \ e^{t^2} \operatorname{erf}(\alpha t + \beta) $$

If $J$ has a 'nice' form, then so does $I$. If that was a $e^{-t^2}$ instead, then it could be done, but even so the result is not pretty and involves the Owen T special function. As an exact solution is not forthcoming, let us investigate asymptotic solutions.

For small $b$ we may expand $e^{bx}$ and integrate term by term. It's useful to note

$$ \frac{e^{bx}}{x}= \sum\limits_{n=0}^\infty \frac{b^n x^{n-1}}{n!} $$

$$ \int\limits_c^\infty dx \ e^{-x^2} x^n =\frac{1}{2} \Gamma \left( \frac{n+1}{2} ,c^2 \right) $$

Where $\Gamma$ is the incomplete gamma function. Thus all the coefficients in the series are known!

$$ I(b)=\frac{1}{2} \sum\limits_{n=0}^\infty \frac{b^n}{n!} \Gamma(n/2,c^2) $$

This series is an exact representation of your integral. We may truncate the series to find an approximation as $b\to 0$. Here is a plot:

For large $b$, we note the integrand has a maxima at $x=b/2$. So long as $c<<b/2$, we may simply do the Gaussian integral (Laplace's method), setting $x=b/2$ in the denominator

$$ I(b) \sim \frac{2\sqrt{\pi}}{b}e^{b^2/4} \ \ , \ b >> 2c $$

For large $c$, integrate by parts

$$ K(c) = \int\limits_c^\infty \frac{dx}{x} e^{-x^2+bx}= \int\limits_c^\infty dx \ \frac{e^{-x^2}}{bx} \frac{d}{dx} \left( e^{bx} \right) \\ $$

$$ K(c)=\frac{e^{-x^2 +bx}}{bc} \bigg\vert_c^\infty +\frac{2}{b} \int\limits_c^\infty \frac{dx}{x} e^{-x^2+bx} + \frac{1}{b} \int\limits_c^\infty \frac{dx}{x^3} e^{-x^2+bx} $$

The second term on the right is $K(c)$. For fixed $b$ and large $c$, the third term differs from $K(c)$ by at least $1/c^2$, so we have

$$ K(c)>>\frac{1}{b} \int\limits_c^\infty \frac{dx}{x^3} e^{-x^2+bx} \ \ , \ \ c \to \infty $$

Neglecting this term, solving for $K$, and simplifying, we find

$$ K(c) \sim \frac{e^{-c^2+bc}}{c(2-b)} \ \ , \ \ c \to \infty $$

Here is a plot:

Finally for $c\to 0$, the best way I've come up with is to use the series representation, split off the $n=0$ term, then use the identities to write

$$ I(c) \sim -\frac{1}{2} \operatorname{Ei}(-c^2) + \sum\limits_{n=1}^N \left[ \frac{b^n \Gamma(n/2)}{2 \cdot n!} -\frac{b^n c^n}{n!n} \right] \ \ , \ \ c\to 0 \text{ and } N \to \infty $$

Where $\operatorname{Ei}$ is the exponential integral, and we have the 'complete' gamma function $\Gamma$. At leading order we have

$$ I(c) \sim -\frac{1}{2} \operatorname{Ei}(-c^2) + b \left( \frac{\sqrt{\pi}}{2}-c \right) \ \ , \ \ c\to 0 \text{ and } b \to 0 $$

Here is a plot: