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I am trying to show that $erf(\sqrt x)=1/\sqrt\pi$ (integral from $t$ to $0$) of $e^{-t}/\sqrt t dt$

I have used the definition of $erf(x) = 2/\sqrt\pi$ (integral from $t$ to $0$) $e^{-t^2} dt$ and used a $u$-substitution and get the answer that I'm looking for in terms of $u$ (which in my substitution is $=x^2$). Is this right? Is it ok that my answer is not converted back to $x$. I'm not really experienced with the error function and am not sure if I'm missing something...

Kenta S
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mayalarson
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  • The upper limit of the integral has to be in terms of x. This has nothing to do with the particular integrand. – herb steinberg Oct 10 '19 at 02:47
  • If I understand your question correctly then you want to show that

    $$\frac{1}{\sqrt{\pi}}\int_0^x\frac{e^{-t}}{\sqrt{t}},dt=\text{erf}(\sqrt{x})$$

    starting from

    $$\text{erf}(\sqrt{x})=\frac{2}{\sqrt{\pi}}\int_0^{\sqrt{x}} e^{-t^2},dt$$

    make the substitution $u=t^2$. Then, $du=2t,dt$ and the integral becomes

    $$\frac{1}{\sqrt{\pi}}\int_0^{\sqrt{x}} \frac{e^{-u}}{\sqrt{u}}du$$

    although I'm not sure what to do with the $\sqrt{x}$. According to Mathematica, the upper limit should be $x$ and not $\sqrt{x}$, so I'm not sure what I did wrong.

     – Axion004 Oct 10 '19 at 05:32

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