Calculate $\frac{d}{dx} \int_{0}^{x} \frac{1}{\sqrt (x - t)} dt$
I want to use Leibnitz rule to calculate this value. But I am unable to find it as the second term would be undefined. Anyone can consider $t = x sin ^2 z$ and then after integrating the integrand, use the differentiation tecnique. It will work. But I want this by using Leibnitz technique. Kindly let me know whether it is possible or not.
$\displaystyle\int_{0}^{x}(x-t)^{-1/2}dt=\lim_{u\rightarrow x}\int_{0}^{u}(x-t)^{-1/2}dt=\lim_{u\rightarrow x}-(1/2)(x-t)^{1/2}\bigg|_{t=0}^{t=x}=(1/2)x^{1/2}$, and then doing differentiation is no problem.
Leibniz rule still goes through, but needs a proof. The idea is that, the integrand is nonnegative, so improper integrable and Lebesgue integrable are the same, by running the classical proof of Leibniz rule, one uses Lebesgue Dominated Convergence Theorem.
Let $F(x,t)$ to be the integrand, note that $t$ is defined only for $t<x$. Now let $G(x)=\displaystyle\int_{0}^{x}F(x,t)dt$, we are to consider $\dfrac{1}{-h}[G(x-h)-G(x)]$ for small $h>0$. Note that we cannot consider about the right derivative, because $F(x,t)$ is not defined for $t\in(x,x+h)$. Now \begin{align*} \dfrac{1}{-h}[G(x-h)-G(x)]=\frac{1}{-h}\left\{\int_{0}^{x-h}F(x-h,t)-F(x,t)dt\right\}+\dfrac{1}{-h}\left\{\int_{x-h}^{x}F(x,t)dt\right\}, \end{align*} now the partial derivative of $F(x,t)$ with respect to $x$ is locally bounded for $x$, and the interval $[0,x]$ is bounded, Lebesgue Dominated Convergence Theorem applies.