# Can $\int_0^{\infty} e^{-a^2 t^2 - b t} \sin c t \mathrm dt$ be done?

I'm a physicist and I've been encountering integrals like $\int_0^{\infty} e^{-a^2 t^2 - b t} \sin c t \;\mathrm dt$, where everything is real.

Mathematica could not solve it and I could not find it in any references. Any idea if this can be solved? How?

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Write $\sin(ct) = \frac{e^{ict} - e^{-ict}}{2i}$ and then the complete the squares and use the good old formula for the integral of $e^{-x^2}$ – user17762 Aug 14 '11 at 4:25
yep, that does it. thanks. not sure why Mathematica couldn't think to convert to exponential form. – BeauGeste Aug 14 '11 at 4:41
I tried it out in wolfram alpha and as you said for some reason, it doesn't compute it. – user17762 Aug 14 '11 at 4:47
Actually Mathematica can do it. With the following input Integrate[Exp[(-a^2)*t^2 - bt]*Sin[ct], {t, 0, Infinity}, Assumptions -> a^2 > 0 && Element[{b, c}, Reals]] The result returned reads $\frac{i \sqrt{\pi } e^{\frac{(b-i c)^2}{4 a^2}} \left(e^{\frac{i b c}{a^2}} \left(|a|-a \text{erf}\left(\frac{b+i c}{2 a}\right)\right)-|a|+a \text{erf}\left(\frac{b-i c}{2 a}\right)\right)}{4 a^2}$. A version 8.0 was used. – Sasha Aug 14 '11 at 4:55
Sasha's comment above shows that it cannot be done in elementary terms. The answer below computes the integral from $-\infty$ to $\infty$ which is a completely different matter. – Christian Blatter Aug 14 '11 at 10:06

EDIT: I originally did the integral over $(-\infty,\infty)$ and got an elementary solution, instead of over the interval $[0,\infty)$ where we get special functions. This has been amended.

Define $I(a,b) = \int_0^{+\infty} \exp(-a t^2-bt)dt$ for $a\in \mathbb{R}_+$ and $b\in\mathbb{C}$. If we complete the square, we get

$$I(a,b) = \int_0^{+\infty} \exp\left[ -a\left(t+\frac{b}{2a}\right)^2+\frac{b^2}{4a} \right]dt$$

$$=\exp\left(\frac{b^2}{4a}\right) \int_0^{+\infty} \exp\left[ -a\left(t+\frac{b}{2a}\right)^2\right]dt$$

$$=\exp\left(\frac{b^2}{4a}\right)\int_{b/2a}^\infty e^{-au^2}du$$

$$=\frac{1}{2} \sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right) \mathrm{erfc}\left(\frac{b}{2\sqrt{a}}\right).$$

In the above, $\mathrm{erfc}(\cdot)$ denotes the complementary error function (which is defined for complex numbers). Now notice that the sine in the original integral can be split into complex exponentials, giving

$$(*)=\frac{I(a^2,b-ci)-I(a^2,b+ci)}{2i}.$$

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This (and some indications in the comments based on the same step) takes for granted that the integral of $\exp(-(t+z)^2)$ over $t$ on the real line does not depend on the complex number $z$. If $z$ were real this fact would follow from a simple change of variable but if $z$ is complex not real, one cannot help to wonder whether the general theorems from complex analysis and the specific properties of the function $z\mapsto\exp(-z^2)$ on which this is based are really that clear in the author's mind... Maybe they are. – Did Aug 14 '11 at 7:09
@Didier: You're perceptive to notice how I skimped out on explaining that maneuver. :) It's my understanding that this works because $\exp(-z^2)$ is entire and the asymptotic direction of the contour remain preserved, but I don't know the specific theorems which allow for this, as it's only something I've picked up from casual reading. – anon Aug 14 '11 at 7:30
Yes: the integral of any entire function on a loop is zero. Fix a positive real number $y$ and see what happens for your function on the clockwise boundary of the rectangle $(-R,+R)\times(0,y)$. The bottom side yields minus the integral of $\exp(-t^2)$ from $t=-R$ to $t=R$. The top side yields the integral of $\exp(-(t+\text{i}y)^2)$ from $t=-R$ to $t=R$. When $R\to+\infty$, the integrals along both vertical sides go to zero (prove this!) and the integrals along the horizontal sides go to integrals over $t$ on the real line. For every $R$, the sum of these four integrals is zero. Done. – Did Aug 14 '11 at 7:47
Does the integral over $(-\infty,+\infty)$ help with the integral over $[0,+\infty)$? – robjohn Aug 14 '11 at 13:24
I echo @robjohn does the integral anon computed relate to the original one? The integrand is not even as far as I can tell. – BeauGeste Aug 14 '11 at 19:13

This is the imaginary part of $J(a,z)=\displaystyle\int_0^{+\infty}\exp(-a^2t^2+zt)\mathrm{d}t$ for $z=-b+\mathrm{i}c$ hence let us compute $J(a,z)$ for every nonzero real number $a$ and every complex number $z$.

Expanding $\mathrm{e}^{zt}$ as a series in $t$, one gets $$J(a,z)=\sum_{n\ge0}\frac{z^n}{n!}\int_0^{+\infty}\exp(-a^2t^2)t^n\mathrm{d}t.$$ The change of variable $u=a^2t^2$ shows that each integral in the RHS is a multiple of the function $\Gamma$ evaluated at $\frac12(n+1)$, more precisely, $$J(a,z)=\frac1{2a}\sum_{n\ge0}j_n\left(\frac{z}a\right)^n,\qquad j_n=\frac1{n!}\Gamma\left(\frac{n+1}2\right).$$ For every nonnegative integer $k$, $$j_{2k}=\frac{\sqrt\pi}{2^{2k}}\frac1{k!},\quad j_{2k+1}=\frac{k!}{(2k+1)!}.$$ Finally, the integral of interest is $$\frac1{4\mathrm{i}}\sum_{n\ge0}(-1)^n\frac{j_n}{a^{n+1}}((b-\mathrm{i}c)^n-(b+\mathrm{i}c)^n).$$

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