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I noticed after evaluating a form of the error function $$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt$$ on WolframAlpha that another integral representation for $x\in \mathbb{R}$ is $$\text{erf}(x)=\frac{2}{\pi}\int_0^\infty\frac{e^{-t^2}\sin(2tx)}{t}dt.$$ My question is how to derive the latter eq from the former (and vise versa). I've tried a series expansion of the second eq, which gets me nothing but a big algebraic mess. Any thoughts/hints would be appreciated! (Note: this is not HW, just a question out of curiosity.)

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  • Write $$\frac{\sin(2xt)}{2t} =\int^x_0 \cos(2ts) ds $$ then swap the range of integration, maybe? Not sure if this will work, just guessing. – User8128 Mar 23 '16 at 05:22

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If you expand $\sin(2tx)$ in a Maclaurin series and then switch the order of integration and summation, you'll end up with the Maclaurin series of the error function.

$$ \begin{align} \frac{2}{\pi} \int_{0}^{\infty} \frac{e^{-t^{2}} \sin(2tx)}{t} &= \frac{2}{\pi} \int_{0}^{\infty} \frac{e^{-t^{2}}}{t} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!} (2tx)^{2n+1} \, dt \\ &= \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n} (2x)^{2n+1}}{(2n+1)!} \int_{0}^{\infty} t^{2n} e^{-t^{2}} \, dt \\ &= \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n} (2x)^{2n+1}}{(2n+1)!} \frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2} \tag{1} \\ &= \frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{n!(2n+1)} \\ &= \text{erf}(x) \tag{2} \end{align}$$

$(1)$ Please show $\int_0^\infty x^{2n} e^{-x^2}\mathrm dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$ without gamma function?

$(2)$ https://en.wikipedia.org/wiki/Error_function#Taylor_series

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