I am stuck on the following problem evaluating an integral with parameters, where the parameter has a limit: $$\lim \limits_{x \to \infty} \int_0^\infty \sin \left(e^{xt}\right)\,dt$$

I know that in some cases you can differentiate what is contained within the integral, which gives us, $\lim \limits_{x \to \infty} \int_0^\infty t e^{xt}\cos \left(e^{xt}\right)\,dt$, which, with a $u$-substitution leads to $\lim \limits_{x \to \infty} \int_1^\infty \cos \left(u\right)\,dt$, but I don't know for sure that I can do that in this case or what to do from there.


Change variables as $u = e^{xt}$.


$$\left|\int_0^{\infty} \sin(e^{xt})\,dt\right| = \left|\frac1{x}\int_1^{\infty}\frac{\sin u}{u}\, du\right|$$

The integral on the RHS converges. Integrating by parts,

$$\left|\int_1^{\infty}\frac{\sin u}{u}\, du\right|= \left|\cos(1) - \int_1^{\infty}\frac{\cos u}{u^2}\, du\right|< \cos(1) + \int_1^{\infty}\frac{1}{u^2}\, du<\infty$$


$$\lim_{x \to \infty}\int_0^{\infty} \sin(e^{xt})\,dt=0$$

  • $\begingroup$ It is an improper Riemann integral and converges in that sense. However, it is not Lebesgue integrable. $\endgroup$ – robjohn Mar 9 '15 at 4:46
  • $\begingroup$ @robjohn - yes its not absolutely integrable. But I'm not using that. $\left|\frac1{x}\int_1^{\infty}\frac{\sin u}{u}\, du\right|$ = (finite number)$/|x| \to 0$. $\endgroup$ – RRL Mar 9 '15 at 4:53

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