How to integral $\int_{\sqrt{a}}^{\infty}{ e^{-t^2+\beta t} \sin(\beta t)}~\mathrm{d}t$?

Question

I would like to plot the following integral but it gives undefined. Please kindly help me about solving the following integral to plot :

$$ I(t)=\int_{\sqrt{a}}^{\infty}{ e^{-t^2+\beta t} \sin(\beta t)}~\mathrm{d}t $$

where $a>0$, $\beta>0$. Thanks!

Answer 1

Possible solution:

Start by completing the square in the exponential part:

$$-t^2 + \beta t = -\left(t - \frac{\beta}{2}\right)^2 + \frac{\beta^2}{4}$$

So you have for the moment:

$$I(t) = \int_{\sqrt{\alpha}}^{+\infty} e^{-\left(t - \frac{\beta}{2}\right)^2 + \frac{\beta}{4}} \sin(\beta t)\ \text{d}t$$

Now operate the shift

$$t - \frac{\beta}{2} = y$$ so that $\text{d}y = \text{d}t$ and check the extrema of the integral. Taking out the exponential that does not depends upon $t$ and for the moment you got:

$$I(t) = e^{\frac{\beta^2}{4}}\ \int_{\sqrt{\alpha} - \frac{\beta}{2}}^{+\infty} e^{-y^2}\sin\left(\beta y - \frac{\beta^2}{2}\right)\ \text{d}y$$

At this point for the integration in $\text{d}y$ just use the exponential Sine form:

$$\sin\left(\beta y - \frac{\beta^2}{2}\right) \equiv \frac{e^{i\left(\beta y - \frac{\beta^2}{2}\right)} - e^{i\left(\beta y - \frac{\beta^2}{2}\right)}}{2i}$$

You obtain then:

$$I(t) = e^{\frac{\beta^2}{4}}\int_{\sqrt{\alpha}-\frac{\beta}{2}}^{+\infty} e^{-y^2}\left(\dfrac{e^{i\beta y - i\frac{\beta^2}{2}} - e^{-i\beta y + i\frac{\beta^2}{2}}}{2i}\right) \text{d} y$$

Now just take off to the integral those exponential in $\beta$ without any dependence upon $y$ and re-arrange the remaining terms. You will obtain:

(Split the two integrals)

$$I(t) = \dfrac{1}{2i}e^{\frac{\beta^2}{4}}e^{-i\frac{\beta^2}{2}}\int_{\sqrt{\alpha}-\frac{\beta}{2}}^{+\infty} e^{-y(y - i\beta)}\text{d} y - \dfrac{1}{2i}e^{\frac{\beta^2}{4}}e^{i\frac{\beta^2}{2}}\int_{\sqrt{\alpha}-\frac{\beta}{2}}^{+\infty} e^{-y(y + i\beta)}\text{d}y $$

To calculate them, we have another time to use the trick of the square completing: In the first integral we have:

$$-y^2 + i\beta y \equiv -\left(y - \frac{i\beta}{2}\right)^2 - \frac{\beta^2}{4}$$

and in the second one we have:

$$-y^2 - i\beta y \equiv -\left(y + \frac{i\beta}{2}\right)^2 - \frac{\beta^2}{4}$$

Use them, substitute and take out the independent part of the exponential and you will get:

$$I(t) = \dfrac{1}{2i}e^{\frac{\beta^2}{4}}e^{-\frac{\beta^2}{4}}\left[e^{-\frac{i\beta^2}{2}}\int_{\sqrt{\alpha}-\frac{\beta}{2}}^{+\infty}e^{-\left(y - \frac{i\beta}{2}\right)^2}\text{d}y - e^{\frac{i\beta^2}{2}}\int_{\sqrt{\alpha}-\frac{\beta}{2}}^{+\infty}e^{-\left(y + \frac{i\beta}{2}\right)^2}\text{d}y \right]$$

Now, if you are familiar with Special Functions you may know this fundamental special integral function:

$$\int_{a}^{+\infty} e^{-(x\pm b)^2}\text{d}x = \frac{\sqrt{\pi}}{2}\text{Erfc}[a \pm b]$$

Using this relation and you will easily integrate the two terms above, obtaining in the end:

$$ \boxed{I(t) = \dfrac{1}{2i}\left[e^{-\frac{i\beta^2}{2}} \dfrac{\sqrt{\pi}}{2}\text{Erfc}\left(\sqrt{\alpha} - \dfrac{\beta}{2} + i\dfrac{\beta}{2}\right) - e^{\frac{i\beta^2}{2}} \dfrac{\sqrt{\pi}}{2}\text{Erfc}\left(\sqrt{\alpha} - \dfrac{\beta}{2} - i\dfrac{\beta}{2}\right)\right]} $$

Honestly, this result is clearly correct, modulo some signs errors or constants missed. Check the whole procedure please, because it is the way to come to the result, but in the very last passages there could be some missing constants.

Answer 2

Explaining David G. Stork's result, $$ \begin{aligned} I &=\int_{\sqrt a}^\infty e^{-t^2+\beta t} \sin(\beta t) \, dt \\ &= \Im \, \int_{\sqrt a}^\infty e^{-t^2+\beta t + i\beta t} \, dt \\ &= \Im \left[ e^{(\beta + i\beta)^2/4} \int_{\sqrt a - \beta(1+i)/2}^\infty e^{-u^2} \, du \right] \\ &= \Im \left[ \frac{\sqrt\pi}{2} \, e^{i\beta^2/2} \, \mathrm{erfc}\left(\sqrt a - \frac{\beta(1+i)}{2}\right) \right] \\ &= \frac{\sqrt\pi}{4i} \left[ e^{i\beta^2/2} \, \mathrm{erfc}\left(\sqrt a - \frac{\beta(1+i)}{2}\right) - e^{-i\beta^2/2} \, \mathrm{erfc}\left(\sqrt a - \frac{\beta(1-i)}{2}\right) \right]. \end{aligned} $$

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Edit. It appears to be hard to get rid of the complex numbers. Practically, we make a series expansion: $$ \mathrm{erfc}(x + yi) =\mathrm{erfc}(x) +\sum_{k = 1}^\infty \frac{2(-i)^k y^k}{\sqrt\pi k!} H_{k-1}(x) e^{-x^2}, $$ where $H_{k-1}(x)$ is the physicists' Hermite polynomial. The series expansion, with only real numbers is $$ \begin{aligned} I &=\frac{\sqrt\pi}{2}\mathrm{erfc}(x)\sin\frac{\beta^2}{2}+e^{-x^2} \sum_{k = 1}^\infty (-)^k \left[ \frac{\beta^{2k} H_{2k-1}(x)}{2^{2k} (2k)!} \sin\frac{\beta^2}{2} - \frac{\beta^{2k-1} H_{2k-2}(x)}{2^{2k-1} (2k-1)!} \cos\frac{\beta^2}{2}, \right] \end{aligned} $$ where $x = \sqrt{a} -\beta/2$.