Chemistry - physical chemistry assignment 1 | Physical chemistry homework help

1. For this question you are going to solve this identity that is very frequently

encountered in PChem II:

∫ x2e−ax 2 ∙ ∂x c = √π 4 ∙ a3 2⁄ ⋅ erf(√a ∙ c) − c 2a e−a∙c

where “erf” is the “error function”. To do so, please use the following steps:

a) Start with substituting y2 = ax2, and don’t forget to change the upper and lower limits!

Here is the formula for variable substitution for x → y, where g(y) = x:

∫ f(x) ∙ ∂x c → ∫ f(g(y)) ∙ | ∂g(y) ∂y | ∂y g(y)=c g(y)=0 b) Starting with 1 √a ∫ y2 ∙ e−y 2 ∙ ∂y √a∙c

0 use integration by parts:

∫f(y) ∙ ∂g(y) c = f(y) ∙ g(y)|0 c −∫g(y) ∙ ∂f(y) c and don’t forget to apply limits on the 1st term on the right. The integral in the 2nd term

on the right will be solved using this identity:

∫ e−x 2 ∙ ∂x c = 1 2 √π ∙ erf(c) where “erf” is just a function like sine or tangent etc.