# Thread: Differentiation under the integral sign

1. ## Differentiation under the integral sign

Ok, could someone please explain this to me, for I have tried to find it on the internet but everywhere (especially the wikipedia) article is written from an analysis point of view. Functionally I would like to know how it works.

I know that

If $\displaystyle I(\theta)=\int_a^{b}f(x,\theta)d\theta$

That by Lebniz

$\displaystyle \frac{dI(\theta)}{d\theta}=\int_a^{b}f(x,\theta)dx$

So now here is where I am in the dark, does this imply that lets say we integrate the right hand side and get

$\displaystyle F(\theta)$ (obviously there will be no more x's)[/tex]

Then we have that

$\displaystyle I(\theta)=\int{F(\theta)}d\theta$

So

Like

$\displaystyle \int_a^{b}f(x,5)dx=\int{F(\theta)}d\theta\bigg|_{\ theta=5}$

Please if anyone could enlighten me, I have been curious about this method for a while and cannot seem to find a comprehensible source. So if somone could explain it that would be great.

Mathstud

2. $\displaystyle \int_a^b f(x,\theta)~d\theta$ is a function of x, not $\displaystyle \theta$.

3. Originally Posted by wingless
$\displaystyle \int_a^b f(x,\theta)~d\theta$ is a function of x, not $\displaystyle \theta$.
If you are talking about my first integral that was a typo it should have been dx.

4. I don't think I understand what you are asking.

If $\displaystyle I(\theta)=\int_a^{b}f(x,\theta)~d\theta$,

then $\displaystyle \frac{dI(\theta)}{d\theta}=\int_a^{b}\frac{\partia l}{\partial \theta}f(x,\theta)~dx$

Can you fix the typos and ask it more clearly?

5. Originally Posted by wingless
I don't think I understand what you are asking.

If $\displaystyle I(\theta)=\int_a^{b}f(x,\theta)~d\theta$,

then $\displaystyle \frac{dI(\theta)}{d\theta}=\int_a^{b}\frac{\partia l}{\partial \theta}f(x,\theta)~dx$

Can you fix the typos and ask it more clearly?

So say you have an an integral and you set it up so it looks like this

$\displaystyle I(\theta)=\int_a^{b}f(x,\theta)dx$

where theta is a fixed constant

I believe that

$\displaystyle \frac{dI(\theta)}{d\theta}=\int_a^b\frac{\partial{ f(x,theta)}}{\partial{\theta}}dx$

So now if we inegrated the right side we would get a function of theta

So lets call that function $\displaystyle F(\theta)$

So now we have that

$\displaystyle \frac{dI(\theta)}{d\theta}=F(\theta)$

But since $\displaystyle I(\theta)$ was our original quantity we wanted to find

I suppose (and this is where I am unsure)

That you then compute $\displaystyle \int{F(\theta)}d\theta$

Because

$\displaystyle I(\theta)=\int{F(\theta)}d\theta$

If so, where does the constant of integration go?