Evaluating $\int_0^{\infty}\!\sin(ax)\,dx$ I've been told by somebody (who is more advanced in math than me) that
$$
\int_0^{\infty}\!\sin(ax)\,dx = \frac{1}{a}
$$
But my basic intuition that the integral is the area under the curve $\sin(ax)$ makes me feel that it could have any value between $0$ and $1$. When questioned further she told me that one could do it by Laplace transforms. Although I don't exactly know what Laplace transforms are, or how they can be used to evaluate integrals, isn't my intuition more correct?
Thank you!
 A: Your intuition is "more correct", though your range $[0,1]$ is not usually correct. You have  $$\int_0^t \sin(ax)\;dx =  \frac{1-\cos\left( a\,t\right) }{a}$$ which, depending on $t$ can take values in the range $\left[0, \frac{2}{a} \right]$, and so does not converge.
Using the Wikipedia article on the Laplace transform, using $$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt$$ then with $f(t)= \sin(\omega t) \cdot u(t)$ [where $u(t)=1$ for $t \ge 0$], you get $$F(s) = \mathcal{L} \left\{f(t)\right\}= \frac{\omega}{s^2+\omega^2}.$$
If you were to let $s=0$ and $\omega = a$ in this last expression, then you would indeed get  $\int_0^{\infty} \sin(ax)\;dx =\frac{1}{a}$ as you were told.  But as the Wikipedia article states, there is only convergence when the real part of $s$ is positive, i.e. not when $s=0$. 
A: $$\int_0^\infty \sin(ax) dx = \lim_{M \to \infty} \int_0^M \sin(ax) dx = \lim_{M \to \infty} \frac{1}{a}\left(1-\cos(aM)\right)$$
which oscilates between 0 and $2/a$ - hence the limit actually doesn't exist. I don't know under which sense that can be say to have a limit $1/a$. As commented by Corey, in the Riemann sense this integral is not convergent. And it's not Lebesgue integrable either.
