# The $\lim_{t\to \infty}\frac{\int_0^\infty \sin(r)r\,dr}{te^t}$ doesn't exist?

We have that $\int_0^\infty \sin(r)r \, dr=\pm \infty$ (doesn't exist) so my guess is that

$\lim_{t\to \infty}\frac{\int_0^\infty \sin(r)r\,dr}{te^t}=\pm \infty$ even though the exponential grows faster.

We can also write this as $\lim_{t\to \infty}\lim_{R\to \infty}\frac{\int_0^R \sin(r)r\,dr}{te^t}$.

Thanks

• Did you mean for the integral to have bounds which depend on $t$? – JimmyK4542 Jan 2 '15 at 2:15
• no, I meant what I wrote. This is not a textbook question. – TKM Jan 2 '15 at 2:58
• Perhaps they meant $\displaystyle\lim_{t\to\infty}\frac{\displaystyle\int_0^t\sin(r)~r~dr}{t~e^t}$ ? – Lucian Jan 2 '15 at 12:25

Since $I=\int_0^{\infty}r\sin r\,dr$ doesn't exist, it follows that there is no real number $t$ for which $I/(te^t)$ exists, and a fortiori $\lim_{t\to\infty}(I/(te^t))$ can't possibly exist.