Does $I(x)$ exists and is it continous on $[0,\infty)$?

Question

Let $f(x,t)=xe^{-xt}$. Show that the integral $I(x)=\int^{\infty}_0f(x,t)dt$ exists for all $x\geq0$. Is $x\to I(x)$ continuous on $[0,\infty)$?

My attempt:

$\int^{\infty}_0xe^{-xt}dt=$ Substitution $u=-xt$ and then $du=-xdx$ so the integral is:

$$x\int^0_1\frac{e^u}{-x}=\int^1_0e^udu=e-1$$

So, since $I(x)$ is constant on $x\geq 0$, $I(x)$ exists and is continuous on $[0,\infty)$

Is this correct? Any help is appreciated

Answer 1

It's almost correct. Note that if you substitute $ u = - x t$ then the integral extrema remain $0$ and $ +\infty$.

My approach:

First we note that $I(0)=0$ (why?), then since $x$ does not depend of $t$ you can take it out of the integral obtaining $$ I(x) = \int_0^{+\infty} x \operatorname{e}^{-xt} \operatorname d t = x\int_0^{+\infty} \operatorname{e}^{-xt} \operatorname d t $$ And now $$ I(x) = x \frac{\operatorname{e}^{-xt} }{-x} \Bigg|^{+\infty}_0= -\operatorname{e}^{-xt}\bigg|_0^{+\infty} = 0 +1 = 1.$$ So we get $$ I(x) = \begin{cases} 0 & \text{if $x = 0$,} \\ 1 & \text{elsewhere}. \end{cases} $$ So $I$ exists but it isn't continous.

Observe that I used the fact that $x \ge 0$. Where?