# Differentiating Under Integral Sign

1. Feb 8, 2016

### iggyonphysics

I(α) = 0∫e-(x2+α/x2) dx

Differentiating under the integral sign leads to:

I(α) = 0∫-e-(x2+α/x2)/x2 dx

Here I am supposed to let u = sqrt(a)/x, but the -x2 doesn't cancel out,

Wolfram-Alpha tells me the answer is: e(-2 sqrt(α) sqrt(π))/(2 sqrt(α)). I understand where the sqrt(π))/(2)sqrt(α) comes from, but not the 2 sqrt(α)) in the numerator.

Thanks!

Last edited: Feb 8, 2016
2. Feb 8, 2016

### Staff: Mentor

Differentiating what with respect to what, and why?
Your integral is missing something ("dx"?).

3. Feb 8, 2016

### HallsofIvy

Staff Emeritus
Yes, the derivative, with respect to "a", of $\int_0^\infty e^{-x^2- a/x^2}dx$ is $-\int_0^\infty \frac{e^{-x^2- a/x^3}}{x^2}dx$.

Is your question about actually doing that integration? Are you required to?

4. Feb 8, 2016

### iggyonphysics

Yes, with dx. Sorry! (Also, how do you format equations?)

For this problem, I am supposed to find I(1) using u = sqrt(x)/a and the answer is e-2sqrt(π)/2 .

5. Feb 9, 2016

### Samy_A

You have $\displaystyle l(a)=\int_0^\infty e^{-x^2- a/x^2}dx$
Then $\displaystyle l'(a)=-\int_0^\infty \frac{e^{-x^2- a/x^2}}{x^2}dx$
Apply the substitution $u=\frac{\sqrt a}{x}$ to this last integral, and compare what you get with the expression for $l(a)$.
This will give you an easy differential equation for $l$.

You will have to use the value of the Gaussian integral at some point:
$\displaystyle l(0)=\int_0^\infty e^{-x^2}dx=\frac{\sqrt \pi}{2}$.

See the LaTeX link at the bottom left of the edit box.

Last edited: Feb 9, 2016
6. Feb 9, 2016

### iggyonphysics

Right, so I have $I'(α) = 1/\sqrt{α}\int_0^∞ e^{-x^2 - u^2} du = 1/\sqrt{α}\int_0^∞ e^{-a/u^2 - u^2} du$, which looks just like the $l(a)=\int_0^\infty e^{-x^2- a/x^2}dx$

I don't understand what to do next.

7. Feb 9, 2016

### Samy_A

I also have a minus sign, giving:

$\displaystyle I'(\alpha)=-\frac{1}{\sqrt \alpha}\int_0^\infty e^{-x^2- \alpha /x^2}dx=-\frac{1}{\sqrt \alpha}I(\alpha)$

Now solve the differential equation $I'(\alpha)=-\frac{1}{\sqrt \alpha}I(\alpha)$.

Last edited: Feb 9, 2016
8. Feb 9, 2016

### iggyonphysics

Got it, thanks so much!