R(x) := ∫ exp ( -y^2 - x^2/y^2 ) dy
The Attempt at a Solution
I move the derivative operator inside the integral and differentiate with respect to x
R'(x) = ∫ [ - 2x/y^2 ] exp ( -x^2/y^2 - y^2 ) dy
Then I let: t = x/y and dy = - x/t^2 dt
R'(x) = 2 ∫ [ - x ] [ t^2 / x^2 ] exp ( t^2 - x^2/t^2 ) [ - x/t^2 ] dt
=> R'(x) = 2 R(x)
But that last part is supposed to be R'(x) = - 2 R(x) - I don't see why.
Then, it follows that this integrates to R(x) = Ae^(−2x) and x = 0 gives R(0) = √π. This last part I don't get it neither. From where the e^(-2x) came from?
Any help is very appreciated.