find the derivative of the integral

Question

Prove that the following integral $F(x)$ is differentiable for every $x \in \mathbb{R}$ and calculate its derivative.

$$F(x) = \int\limits_0^1 e^{|x-y|} \mathrm{d}y$$

I don't know how to get rid of the absolute value in the integral

any ideas?

Answer 1

Split the integral into different cases.

1. If $x\geq1$, we have $$F(x) = \int_0^1 e^{x-y}dy = e^x \int_0^1 e^{-y}dy = e^x\left(1-e^{-1}\right)$$
2. If $x\leq0$, we have $$F(x) = \int_0^1 e^{y-x}dy = e^{-x} \int_0^1 e^{y}dy = e^x\left(e-1\right)$$
3. If $x \in (0,1)$, we have $$F(x) = \int_0^x e^{x-y}dy + \int_x^1 e^{y-x}dy = e^x\left(1-e^{-x}\right) + e^{-x}\left(e-e^x\right) = e^x+ e^{1-x} - 2$$

Answer 2

hint

assume you are integrating $dy$ not $dx$ as written.

consider 3 cases with $x<0, x>1$ and otherwise, and in the 3rd case, break the integral in two.

Answer 3

Let $\phi_y(x) = e^{|x-y|}$.

The mean value theorem shows that $|e^a-e^b| \le e^{\max(a,b)} |a-b|$, and also, $||a|-|b|| \le |a-b|$, hence we have $|{\phi_y(x+h)-\phi_y(x) \over h}| \le e^{\max(|x-y|,|x+h-y|)} |h|$ for $h \neq 0$.

If $\phi_y(x) = e^{|x-y|}$ then $\phi_y'(x) = e^{|x-y|}$ for $x>y$ and $\phi_y'(x) = -e^{|x-y|}$ for $x <y$.

If we fix $x$ and take $|h| < 1$, then we have $e^{\max(|x-y|,|x+h-y|)} \le M$ for some $M$ and all $y \in [0,1]$, hence the quotient ${\phi_y(x+h)-\phi_y(x) \over h}$ is uniformly bounded by $M$ and ${\phi_y(x+h)-\phi_y(x) \over h} \to \phi_y'(x)$ for ae. $y$. The dominated convergence theorem gives $F'(x) = \int_0^1 \phi_y'(x) dy$, hence we have $F'(x) = \int_0^1 \operatorname{sgn}(x-y)\phi_y(x) dy = \int_0^1 \operatorname{sgn}(x-y)e^{|x-y|} dy$.

Answer 4

$\dfrac{dF(x)}{dx}$

$=\dfrac{d}{dx}\int_0^1e^{|x-y|}~dy$

$=\dfrac{d}{dx}\int_0^1e^{(x-y)\text{sgn}(x-y)}~dy$

$=\int_0^1\text{sgn}(x-y)e^{(x-y)\text{sgn}(x-y)}~dy$

$=[-e^{(x-y)\text{sgn}(x-y)}]_0^1$

$=[-e^{|x-y|}]_0^1$

$=e^{|x|}-e^{|x-1|}$