Benutzer:Dirk Huenniger/heaviside

Just for fun I calculate the convolution of a heaviside function with a gaussian. The convolution is defined by:

$(f*g)(t)=\int _{-\infty }^{\infty }f(u)g(t-u)\mathrm {d} u$

Let $H$ denote the Heaviside function.

$(f*H)(t)=\int _{-\infty }^{\infty }f(u)H(t-u)\mathrm {d} u=\int _{-\infty }^{t}f(u)\mathrm {d} u$

Let f be a Gaussian $f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}\mathrm {exp} \left({-{\frac {x^{2}}{2\sigma ^{2}}}}\right)=0$

So we got $(f*H)(t)=\int _{-\infty }^{\infty }f(u)H(t-u)\mathrm {d} u={\frac {1}{\sigma {\sqrt {2\pi }}}}\int _{-\infty }^{t}\mathrm {exp} \left({-{\frac {u^{2}}{2\sigma ^{2}}}}\right)\mathrm {d} u$

Since

$\operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}dt.$

Let us substitute

$z={\frac {1}{{\sqrt {2}}\sigma }}u$

So

${\frac {\mathrm {d} z}{\mathrm {d} u}}={\frac {1}{{\sqrt {2}}\sigma }}$

and

${\frac {\mathrm {d} u}{\mathrm {d} z}}={\sqrt {2}}\sigma$

We got

$(f*H)(t)={\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{{\frac {1}{{\sqrt {2}}\sigma }}t}\mathrm {exp} \left(-z^{2}\right)\mathrm {d} z={\frac {1}{2}}\operatorname {erf} ({\frac {1}{{\sqrt {2}}\sigma }}t)$

So we see that the information about the error $\sigma$ has survived the transfirmation. Which is what I was after.