# Differentiation under the integral sign

## Homework Statement

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.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

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.

I also get R'(x) = 2*[R(x) + c], where c is a constant of integration. Here are the steps, done in Maple 11 (with output written in LaTeX):

R:=Int(exp(-x^2/y^2-y^2),y);
$$R = \int e^{\left(\displaystyle -\frac{x^2}{y^2} - y^2\right)} \, dy$$
Rp:=diff(R,x);
$$Rp = \int -\frac{2x}{y^2} e^{\left(\displaystyle -\frac{x^2}{y^2} - y^2\right)} \, dy$$
changevar(x/y=t,Rp);
$$\int e^{\left(\displaystyle -\frac{x^2}{t^2} - t^2\right)} \, dt$$

RGV

I just realize that I didn't consider the limits of integration... The limits are from 0 to infinity...

but still something must be wrong...

Yeah, the negative probably comes in from having to flip the limits of integration when you transform to the varaible t.

HallsofIvy
Homework Helper
Does "exp(-y^2- x^2/y^2)" mean $e^{-y^2- (x^2/y^2)}$ or $e^{(-y^2- x^2)/y^2}$.

Hello Halls,

The one in the middle.

SammyS
Staff Emeritus
Homework Helper
Gold Member
I just realize that I didn't consider the limits of integration... The limits are from 0 to infinity...

but still something must be wrong...

Does "exp(-y^2- x^2/y^2)" mean $e^{-y^2- (x^2/y^2)}$ or $e^{(-y^2- x^2)/y^2}$.

So the definition of R(x) is:
$\displaystyle R(x) := \int_{0}^{\infty} e^{\left( -\frac{x^2}{y^2} - y^2\right)} \, dy$​
That explains why R is not a function of y.