Differentiate under integral of function with intervals $-\infty$ and a function of $x$ (can't use Leibniz?) I need to differentiate an integral with respect to $x$:
$$\frac{d}{dx} \int_{-\infty}^{\frac{a(x)}{D}} n(\delta,x)\mathrm d\delta$$ 
I have some result already which I am trying to check or arrive at:
$$\int_{-\infty}^{\frac{a(x)}{D}} \frac{dn(\delta,x)}{dx}\mathrm d\delta +\left. \frac{1}{D}\frac{\mathrm dn(\delta,x)}{\mathrm d\delta}\right|_{\delta=\frac{a(x)}{D}}$$ 
But I cannot see how to arrive at the second term. I believe it is not possible to use Leibniz since the lower limit of the integral goes to minus infinity. Please could you suggest an approach for tackling this problem?
Thanks for reading.
 A: The leibniz integral rule for functions of your notation looks like:
$$\frac{d}{dx} \int_{a(x)}^{b(x)} n(\delta,x) \mathrm{d}\delta = \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} n (\delta,x) \mathrm{d}\delta + n\left(b(x),x\right) \frac{d}{dx} b(x) - n(a(x),x) \frac{d}{dx}a(x)$$
in your specific case we have, where $c\in\mathbb{R}$,
\begin{align}
\frac{d}{dx}\lim_{c\to-\infty} \int_{c}^{\frac{a(x)}{D}} n(\delta,x) \mathrm{d}\delta = &\lim_{c\to - \infty}\int_{c}^{\frac{a(x)}{D}} \frac{\partial}{\partial x} n (\delta,x) \mathrm{d}\delta \\&+ n\left(\frac{a(x)}{D},x\right) \frac{d}{dx} \left(\frac{a(x)}{D}\right){ - \lim_{c\to - \infty}n(c,x) \frac{d}{dx}c}
\end{align}
clearly the last term is $0$ as $c$ is just a real number, so its derivative is $0$. Hence we have,
$$\frac{d}{dx}\lim_{c\to-\infty} \int_{c}^{\frac{a(x)}{D}} n(\delta,x) \mathrm{d}\delta = \lim_{c\to - \infty}\int_{c}^{\frac{a(x)}{D}} \frac{\partial}{\partial x} n (\delta,x) \mathrm{d}\delta + n\left(\frac{a(x)}{D},x\right) \frac{d}{dx} \left(\frac{a(x)}{D}\right)$$
which can be rewritten as:
$$\frac{d}{dx}\lim_{c\to-\infty} \int_{c}^{\frac{a(x)}{D}} n(\delta,x) \mathrm{d}\delta =\lim_{c\to - \infty}\int_{c}^{\frac{a(x)}{D}} \frac{\partial}{\partial x} n (\delta,x) \mathrm{d}\delta +\frac{1}{D} n\left(\frac{a(x)}{D},x\right) \frac{d}{dx}a(x)$$
This is what the answer would look like if you used Leibniz's Integral Rule naturally, but the concern comes with the left hand side of the equation. Let$$  f(c,x) = \int_{c}^{\frac{a(x)}{D}} n(\delta,x) \mathrm{d} \delta $$
As long as
$$\frac{d}{dx}\lim_{c\to-\infty}f(c,x)=\lim_{c\to -\infty}\frac{d}{dx}f(c,x)$$
is true, we are okay. And for that to happen $\lim_{c\to-\infty} f(c,x)$ has to converge for some value of $x$ and $\frac{d}{dx} f(c,x)$ converges as $c\to -\infty$, which only you can really determine because you know the behavior of $n(\delta,x)$
I'm not sure how you have the term $\frac{d}{d\delta} n(\delta,x)$ arising in your claimed solution, because I don't see how a derivative with respect to $\delta$ would arise.
I am not 100% confident of my answer and so I hope others can chime in or give a more rigorous response.
