# Differentiation under the integral sign

1. Jun 8, 2012

### naaa00

1. The problem statement, all variables and given/known data

R(x) := ∫ exp ( -y^2 - x^2/y^2 ) dy

3. 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.

2. Jun 8, 2012

### Ray Vickson

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

3. Jun 8, 2012

### naaa00

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...

4. Jun 8, 2012

### Muphrid

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

5. Jun 8, 2012

### HallsofIvy

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

6. Jun 8, 2012

### naaa00

Hello Halls,

The one in the middle.

7. Jun 8, 2012

### SammyS

Staff Emeritus
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.