How can I show this is equivalent to the error function of $x$?

Question

$$\frac{2}{\pi} \int_0^\infty e^{-t^2}\frac{\sin 2xt}{t}\,dt$$

I know the original $\operatorname{erf}x$ but the infinity in the limit keeps getting in my way. How do I deal with it? Also my hint is that I have to use sine expansion.

Answer 1

METHODOLOGY $1$: Using an integral representation of $\displaystyle \frac{\sin(2xt)}t$

Let $I(x)$ be given by the integral $I(x)=\frac2\pi\int_0^\infty e^{-t^2}\frac{\sin(2xt)}{t}\,dt$. Then, noting that $\frac{\sin(2xt)}t=\int_{-x}^x e^{i2st}\,ds$, we can write

$$\begin{align} I(x)&=\frac2\pi\int_0^\infty e^{-t^2}\frac{\sin(2xt)}{t}\,dt\\\\ &=\frac1\pi\int_{-\infty}^\infty e^{-t^2}\int_{-x}^x e^{i2st}\,ds\,dt\\\\ &=\frac1\pi \int_{-x}^x \int_{-\infty}^\infty e^{-t^2}e^{i2st}\,dt\,ds\\\\ &=\frac1\pi \int_{-x}^x\int_{-\infty}^\infty e^{-s^2}e^{-(t-is)^2}\,dt\,ds\\\\ &=\frac1\pi \int_{-x}^xe^{-s^2}\int_{-\infty-is}^{\infty-is}e^{-t^2}\,dt\,ds\\\\ &=\frac1\pi \int_{-x}^xe^{-s^2}\int_{-\infty}^{\infty}e^{-t^2}\,dt\,ds\\\\ &=\frac{2}{\sqrt \pi}\int_0^x e^{-s^2}\,ds\\\\ &=\text{erf}(x) \end{align}$$

as was to be shown!

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METHODOLOGY $2$: Using an series representation of $\displaystyle \frac{\sin(2xt)}t$

Let $I(x)$ be given by the integral $I(x)=\frac2\pi\int_0^\infty e^{-t^2}\frac{\sin(2xt)}{t}\,dt$. Then, noting that $\frac{\sin(2xt)}t=\sum_{n=0}^\infty \frac{(-1)^n(2xt)^{2n+1}}{(2n+1)!}$, we can write

$$\begin{align} I(x)&=\frac2\pi\int_0^\infty e^{-t^2}\frac{\sin(2xt)}{t}\,dt\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n(2x)^{2n+1}}{(2n+1)!}\int_0^\infty t^{2n}e^{-t2}\,dt\\\\ &=\frac2\pi \sum_{n=0}^\infty \frac{(-1)^n(2x)^{2n+1}}{(2n+1)!}\left(\frac12\frac{(2n)!}{4^n\,n!}\sqrt{\pi}\right)\\\\ &=\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)\,n!}\\\\ &=\frac{2}{\sqrt{\pi}}\text{erf}(x) \end{align}$$

as expected!

Answer 2

Formally, expand $\sin(2xt)$ in its Maclaurin series and integrate term-by-term. To justify the interchange of limits, you can estimate the error in the Maclaurin series, or you can use the Lebesgue dominated convergence theorem, noting that $\int_0^\infty \exp(-t^2 + 2|x|t)\; dt$ converges.