# Thread: what is the value of this integral

1. ## what is the value of this integral

$\int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}}{t^{2}+1}}(t^{2}+1)^{-\frac{1}{2}}dx}$

epsilon is infinitesimal

2. Originally Posted by szpengchao
$\int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}}{t^{2}+1}}(t^{2}+1)^{-\frac{1}{2}}dx}$

epsilon is infinitesimal
Note that $\int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}}{t^{2}+1}}(t^{2}+1)^{-\frac{1}{2}}dx}=\int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\left(\frac{x}{\sqrt{t^{2}+1}}\right)^2}(t^{2}+1)^ {-\frac{1}{2}}dx}$

Let $u=\frac{x}{\sqrt{t^2+1}}\implies \,du=\frac{\,dx}{\sqrt{t^2+1}}$

Thus,

$\int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\left(\frac{x}{\sqrt{t^{2}+1}}\right)^2}(t^{2}+1)^ {-\frac{1}{2}}dx}=\frac{1}{2}\int_{-\frac{\epsilon}{\sqrt{t^2+1}}}^{\frac{\epsilon}{\s qrt{t^2+1}}}\frac{2}{\sqrt{\pi}}e^{-u^2}\,du=\left.\frac{1}{2}\,\text{erf}\left(u\righ t)\right|_{-\frac{\epsilon}{\sqrt{t^2+1}}}^{\frac{\epsilon}{\s qrt{t^2+1}}}$

$=\frac{1}{2}\left[\text{erf}\left(\frac{\epsilon}{\sqrt{t^2+1}}\righ t)-\text{erf}\left(-\frac{\epsilon}{\sqrt{t^2+1}}\right)\right]=\frac{1}{2}\left[2\,\text{erf}\left(\frac{\epsilon}{\sqrt{t^2+1}}\r ight)\right]=\color{red}\boxed{\text{erf}\left(\frac{\epsilon} {\sqrt{t^2+1}}\right)}$

Does this make sense?

--Chris

3. ## what

what is that erf there?

4. Originally Posted by szpengchao
what is that erf there?
Its the error function. See here.