# Thread: Differentiating under the integral sign

1. ## Differentiating under the integral sign

Hello

I know when the limits of integration are finite its possible to bring the derivative sign inside the integral (i.e. the generalized Leibniz integral rule). In its basic form the rule says

$\displaystyle \frac{d}{\text{dx}}\int _{y_0}^{y_1}f(x,y)dy=\int _{y_0}^{y_1}\frac{\partial }{\partial x}f(x,y)dy$.

But what if $\displaystyle y_0 = -\infty$ and $\displaystyle y_1 = + \infty$?

I've seen people bring the derivative sign inside the integral in this case, but since the limits are not finite the Leibniz rule does not apply directly. So how can this operation be performed. What theorem applies??

I have a suspicion it may have something to do with Lebesgues Dominated Convergence Theorem.

2. Originally Posted by aukie
So how can this operation be performed. What theorem applies??
Suppose:

(i) $\displaystyle \int_a^{+\infty}f(x,y)\;dy$ converges forall $\displaystyle x\in [\alpha,\beta]$ . That is the following funtion is well defined:

$\displaystyle F:[\alpha, \beta ]\rightarrow{\mathbb{R}},\quad F(x)= \int_a^{+\infty}f(x,y)dy$

(ii) $\displaystyle \frac{{\partial f}}{{\partial y}}$ is continuous on $\displaystyle [\alpha, \beta ]\times{[a,+\infty)}$

(iii) $\displaystyle \int_a^{+\infty}\frac{{\partial f}}{{\partial x}}\;dy$ uniformly converges on $\displaystyle (\alpha,\beta)$

Then:

(a) There exits $\displaystyle F'(x)$ for all $\displaystyle x\in (\alpha,\beta)$ .

(b) $\displaystyle F'(x)=\int_a^{+\infty}\frac{{\partial f}}{{\partial x}}\;dy$ .

Regards.

Fernando Revilla

3. Hi Fernando, I would like to read more about this, can you direct me towards any references? Uniform convergence is a really strict condition!

4. Originally Posted by aukie
Hi Fernando, I would like to read more about this, can you direct me towards any references? Uniform convergence is a really strict condition!
See for example Tom Apostol Mathematical Analysis, chapter 14.

Regards.

Fernando Revilla