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Term Test 1 / Re: P5
« on: February 16, 2018, 02:50:23 PM »
A direct computation yields:
\begin{align*}
u&=\frac{1}{\sqrt{4t\pi}}\int_{-\infty}^{\infty} \exp{-\frac{(y-x)^2}{4t}-|y|}\,dx\\
&=\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{4t}-y)\,dx +\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{4t}-y)\,dx\\
&=\frac{\exp(x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+2t)^2}{4t})\,dx + \frac{\exp(-x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+2t)^2}{4t})\,dx\\
&=\frac{\exp(x+t)}{\sqrt{\pi}}\int_{\frac{x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz+ \frac{\exp(-x+t)}{\sqrt{\pi}}\int_{\frac{-x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz\\
&=\frac{\exp(x+t)}{2}(1-\text{erf}(\frac{x+2t}{\sqrt{4t}})) + \frac{\exp(-x+t)}{2}(1-\text{erf}(\frac{-x+2t}{\sqrt{4t}}))
\end{align*}
\begin{align*}
u&=\frac{1}{\sqrt{4t\pi}}\int_{-\infty}^{\infty} \exp{-\frac{(y-x)^2}{4t}-|y|}\,dx\\
&=\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{4t}-y)\,dx +\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{4t}-y)\,dx\\
&=\frac{\exp(x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+2t)^2}{4t})\,dx + \frac{\exp(-x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+2t)^2}{4t})\,dx\\
&=\frac{\exp(x+t)}{\sqrt{\pi}}\int_{\frac{x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz+ \frac{\exp(-x+t)}{\sqrt{\pi}}\int_{\frac{-x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz\\
&=\frac{\exp(x+t)}{2}(1-\text{erf}(\frac{x+2t}{\sqrt{4t}})) + \frac{\exp(-x+t)}{2}(1-\text{erf}(\frac{-x+2t}{\sqrt{4t}}))
\end{align*}