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It seems to me that $$e^x=1+\frac{1}{\sqrt{\pi }}{\int_0^x \frac{e^t \text{erf}\left(\sqrt{t}\right)}{\sqrt{x-t}} \, dt}$$ This integral seems to converge for all $x\in\mathbb{C}$

I came upon this conjecture by following the instructions here to do a half integral twice. Can anyone prove this conjecture is true?

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The identity $$ 1=e^{-x}+\frac{1}{\sqrt{\pi }}{\int_0^x \frac{e^{-(x-t)} }{\sqrt{x-t}}\cdot \text{erf}\left(\sqrt{t}\right) \, dt} $$ is true because of Laplace transforms $$ \mathcal{L}\left(1\right)=\frac{1}{p},\qquad \mathcal{L}\left(e^{-t}\right)=\frac{1}{p+1}, $$ $$ \mathcal{L}\left(\frac{e^{-t} }{\sqrt{t}}\right)=\int_0^\infty\frac{e^{-t} }{\sqrt{t}}e^{-pt}\ dt=\frac{1}{\sqrt{p+1}}\int_0^\infty \frac{e^{-t}}{\sqrt{t}}dt=\frac{\sqrt{\pi}}{\sqrt{p+1}}, $$ $$ \mathcal{L}\left( \text{erf}(\sqrt{t}) \right)=\int_0^\infty e^{-pt}\ dt\frac{2}{\sqrt{\pi}}\int_0^\sqrt{t}e^{-x^2}dx=\\\frac{2}{\sqrt{\pi}}\int_0^\infty \frac{e^{-pt}}{p}\frac{d}{dt}\left(\int_0^\sqrt{t}e^{-x^2}dx\right) dt=\\ \frac{2}{\sqrt{\pi}}\int_0^\infty \frac{e^{-pt}}{p}\frac{e^{-t}}{2\sqrt{t}}dt=\frac{1}{\sqrt{\pi}p\sqrt{p+1}}\int_0^\infty \frac{e^{-t}}{\sqrt{t}} dt=\frac{1}{p\sqrt{p+1}} $$ and convolution theorem for Laplace transform $\mathcal{L}\left(\int_0^\tau f(t)g(\tau-t)dt\right)=\mathcal{L}(f(t))\cdot \mathcal{L}(g(t))$: $$ \frac{1}{p}=\frac{1}{p+1}+\frac{1}{\sqrt{\pi }}\cdot\frac{\sqrt{\pi}}{\sqrt{p+1}}\cdot \frac{1}{p\sqrt{p+1}} $$

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  • $\begingroup$ Could you be more specific? I don't know much about Laplace Transforms and am having difficulty following. $\endgroup$ – Nazgand Nov 23 '15 at 2:50
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    $\begingroup$ I think this is the most natural way to prove this identity, but I might be wrong. I'm afraid without some familiarity with Laplace transform one will not be able to follow the proof. $\endgroup$ – Nemo Nov 23 '15 at 9:27
  • $\begingroup$ Now that you mentioned convolution, I can see what is going on. $\endgroup$ – Nazgand Nov 23 '15 at 10:01

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