Solving gaussian integral $\int_1^3 e^{-\frac{t^2}{2}} \,dt$ with polar coordinates

Question

Let's say I have the integral $\displaystyle \int_1^3 e^{-\frac{t^2}{2}} dt$.

Since $\frac{t^2}{2} = ({\frac{t}{\sqrt 2}})^2$ I could thus say $u = \frac{t}{\sqrt 2}$, $\frac{du}{dt} = \frac{1}{\sqrt 2}$, $dt = \sqrt 2 \cdot du$.

This gives me $\sqrt 2 \displaystyle \int_1^3 e^{-u^2} dt$. Since $\displaystyle \int e^{-x^2}dx={\frac{\sqrt\pi}{2}}\text{erf}(x)$ we now have:

$\sqrt 2 \displaystyle \int_1^3 e^{-u^2} dt = \frac{\sqrt\pi}{\sqrt 2}(\text{erf}(\frac{3}{\sqrt 2})- \text{erf}(\frac{1}{\sqrt 2})) = 0.39$.

But how can I solve this integral using the $2\text{D}$-trick with polar coordinates? I know how we come to:

$I^2 = \displaystyle \int_{\mathbb{R}^2} e^{-\frac {x^2+y^2}{2}} dx dy = \int \int r e^{-\frac {r^2}{2}}dr d \varphi$.

But what are the limits of the last integral, the polar coordinates, given we want to calculate our integral $\displaystyle \int_1^3 e^{-\frac{t^2}{2}} dt$?

Answer 1

Using polar coordinates is only easy for -infinity to infinity(r is 0 to infinity and angle is 0 to 2π) or 0 to infinity(ris 0 to infinity and angle is 0 to π/2. For others we will have a square left out for which r is not fixed .