if $f(x)$ is continuously differentiable in $[0,\infty)$, and $\int_{0}^{\infty}f$ and $\int_{0}^{\infty}f'$ converge

Solution 1:

Not sure if the following is simple enough for you. I'll be using the convergence of these well-known integrals: $$I_1=\int_0^{\infty}\frac{\sin(x)}{x}\,dx\qquad I_2=\int_0^{\infty}\cos(x^2)\,dx $$ (They are called Dirichlet integral and Fresnel integral, respectively.) Let $$f(x)=\begin{cases}\dfrac{\sin(x^2)}{x},\quad x\ne0\\0,\quad\quad\qquad x=0\end{cases} $$ Then $$f'(x)=\begin{cases}2\cos(x^2)-\dfrac{\sin(x^2)}{x^2},\quad x\ne0\\1,\qquad\qquad\qquad\quad\qquad x=0\end{cases} $$ You can verify that $f$ and $f'$ are continuous. $\lim_{x\to\infty}f'(x)$ doesn't exist since $\lim_{x\to\infty}\cos(x^2)$ doesn't exist and $\lim_{x\to\infty}\frac{\sin(x^2)}{x^2}=0$. Regarding the convergence, $$\int_0^{\infty}f(x)\,dx=\int_0^{\infty}\frac{\sin(x^2)}{x}\,dx $$ transforms to $\frac 12I_1$ with the substitution $x=\sqrt{t}$. $$\int_0^{\infty}f'(x)\,dx=2\int_0^{\infty}\cos(x^2)\,dx-\int_0^{\infty}\frac{\sin(x^2)}{x^2}$$ The first integral on the RHS is $I_2$. For the other one, integration by parts gives $$\int_0^{\infty}\frac{\sin(x^2)}{x^2}=\sin(x^2)\left(-\frac 1x\right)\bigg\rvert_0^{\infty}+\int_0^{\infty}\frac{2x\cos(x^2)}{x}\,dx\\ =2\int_0^{\infty}\cos(x^2)\,dx$$ so it's convergent as well (and equal to $2I_2$).