What is the derivative of $\int_0^x \frac{f(y)}{\sqrt{x-y} }$?

Question

Suppose I have some integral $$\int_0^x \frac{f(y)}{\sqrt{x-y} }$$ How would I differentiate this with respect to $x$?

The Leibniz rule reads $$\frac{d}{dx} \int_0^x g(x,y) \ dy = g(x,x) +\int_0^x \frac{ \partial }{ \partial x} \frac{f(y)}{\sqrt{x-y}} \ dy$$

for $$g(x,y)=\frac{f(y)}{\sqrt{x-y}}$$

But here by letting $y=x$ I get division by $0$, undefined.

I tried some values of $f(y)$ like for example $f(y)=1$ for which the integral gives $2 \ \sqrt{x} $ whose derivative is defined everywhere but not at $0$.

So how would I differentiate this integral?

Answer 1

I will assume that $f(x)$ is continuously differentiable. Then for $x > 0$,

$$ \int_{0}^{x} \frac{f(y)}{\sqrt{x-y}} \, \mathrm{d}y = \stackrel{(y=xu)}= \int_{0}^{1} \frac{f(xu)\sqrt{x}}{\sqrt{1-u}} \, \mathrm{d}u, $$

and so, differentiating this with respect to $x$ and utilizing the Leibniz's rule gives

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \int_{0}^{x} \frac{f(y)}{\sqrt{x-y}} \, \mathrm{d}y &= \int_{0}^{1} \frac{\partial}{\partial x} \frac{f(xu)\sqrt{x}}{\sqrt{1-u}} \, \mathrm{d}u \tag{$y=xu$} \\ &= \int_{0}^{1} \frac{2uxf'(xu) + f(xu)}{2\sqrt{x}\sqrt{1-u}} \, \mathrm{d}u \\ &= \frac{1}{2x} \int_{0}^{x} \frac{2yf'(y) + f(y)}{\sqrt{x-y}} \, \mathrm{d}y \tag{$y=xu$} \end{align*}

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Remark. The formula

$$\frac{1}{\sqrt{\pi}} \frac{\mathrm{d}}{\mathrm{d}x} \int_{0}^{x} \frac{f(y)}{\sqrt{x-y}} \, \mathrm{d}y$$

is the half-derivative of $f(x)$ in fractional calculus.

Answer 2

Substitution

One thing that comes to mind is to substitute $y\mapsto x-y$ in the integral: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^x\frac{f(y)}{\sqrt{x-y}}\,\mathrm{d}y &=\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x\frac{f(x-y)}{\sqrt{y}}\,\mathrm{d}y\tag{1a}\\ &=\frac{f(0)}{\sqrt{x}}+\int_0^x\frac{f'(x-y)}{\sqrt{y}}\,\mathrm{d}y\tag{1b}\\ &=\frac{f(0)}{\sqrt{x}}+\int_0^x\frac{f'(y)}{\sqrt{x-y}}\,\mathrm{d}y\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: substitute $y\mapsto x-y$
$\text{(1b)}$: differentiate
$\text{(1c)}$: substitute $y\mapsto x-y$

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Limiting a Truncated Integral $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^{x-\epsilon}\frac{f(y)}{\sqrt{x-y}}\,\mathrm{d}y &=\frac{f(x-\epsilon)}{\sqrt\epsilon}-\frac12\int_0^{x-\epsilon}\frac{f(y)}{\sqrt{x-y}^{\,3}}\,\mathrm{d}y\tag{2a}\\ &=\frac{f(x-\epsilon)}{\sqrt\epsilon}-\int_0^{x-\epsilon}f(y)\,\mathrm{d}\frac1{\sqrt{x-y}}\tag{2b}\\ &=\frac{f(x-\epsilon)}{\sqrt\epsilon}-\frac{f(x-\epsilon)}{\sqrt\epsilon}+\frac{f(0)}{\sqrt{x}}+\int_0^{x-\epsilon}\frac{f'(y)}{\sqrt{x-y}}\,\mathrm{d}y\tag{2c}\\ &=\frac{f(0)}{\sqrt{x}}+\int_0^{x-\epsilon}\frac{f'(y)}{\sqrt{x-y}}\,\mathrm{d}y\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: differentiate
$\phantom{\text{(2a):}}$ as $\epsilon\to0^+$, this tends to $\infty-\infty$
$\text{(2b)}$: prepare to integrate by parts
$\text{(2c)}$: integrate by parts
$\text{(2d)}$: cancel

As $\epsilon\to0^+$, $\text{(2d)}$ matches $\text{(1c)}$.