Exponential integral...

Question

I want to solve this integral

$$\int_0^\infty \sin(kx){e^{-ax^2}}dx $$

But I do not want to Exist complex numbers in my answer

for this reason I do not want to Write

$$\sin(kx)=\frac{e^{ikx}-e^{-ikx}}{2i}$$

Any hints on how I should solve it?

Answer 1

$$\int_0^\infty \sin(kx){e^{-ax^2}}dx = {\frac {\sqrt {\pi }}{2\sqrt {a}}{{\rm e}^{-{{{k}^{2}}/({4 a})}}}{\rm erfi} \left({\frac {k}{2\sqrt {a}}} \right) } $$ where the imaginary error function $\mathrm{erfi}$ may be found here .

Answer 2

I should use Taylor's expansion $$Sin(x)=(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)$$ $$\int_0^\infty (x-\frac{x^3}{3!}+\frac{x^5}{5!}){e^{-ax^2}}dx= $$ $$\int_0^\infty (x){e^{-ax^2}}dx -\int_0^\infty \frac{x^3}{3!}{e^{-ax^2}}dx+\int_0^\infty \frac{x^5}{5!}{e^{-ax^2}}dx=$$ $$u=ax^2$$ $$ du=2axdx$$ The first integral $$\frac{1}{2a}\int_0^\infty {e^{-u}}du=\frac{1}{2a}$$ The second integral $$-\frac{1}{3!}\int_0^\infty {u}{e^{-u}}\frac{du}{2a}=-\frac{1}{2a*3!}\Gamma(2)=-\frac{1!}{2a*3!}$$ And third integral: $$\frac{1}{5!}\int_0^\infty \frac{u^2}{a^2}{e^{-u}}\frac{du}{2a}=\frac{1}{2a^3*5!}\Gamma(3)=\frac{2!}{2a*3!}$$

Answer 3

WLOG, let $a=\frac12$, for convenience. We have

$$I_k:=\int_0^\infty \sin(kx){e^{-x^2/2}}dx.$$

Then, differentiating on $k$ and integrating by parts,

$$\frac{dI_k}{dk}=\int_0^\infty x\cos(kx){e^{-x^2/2}}dx=-\left.\cos(kx){e^{-x^2/2}}\right|_0^\infty-k\int_0^\infty \sin(kx){e^{-x^2/2}}dx\\ =1-kI_k.$$

This is a differential equation that we can integrate.

We can rewrite

$$\left(\frac{dI_k}{dk}+kI_k\right)e^{k^2/2}=\frac{d}{dk}\left(I_ke^{k^2/2}\right)=e^{k^2/2},$$

and

$$I_k=(\Psi(k)+C)\,e^{-k^2/2}$$ where $\Psi(k)$ denotes an antiderivative of $e^{k^2/2}$, a scaled version of the imaginary error function.

Using the initial condition $I_0=0$,

$$I_k=(\Psi(k)-\Psi(0))\,e^{-k^2/2}$$