
Intergral
Let
$\displaystyle f(x)= \int_{x}^{x+1} sin(e^t) \, dt$
Show that
$\displaystyle e^xf(x)<2$
and that
$\displaystyle e^x f(x)=cos(e^x)\frac{cos(e^{x+1})}{e}+r(x)$
where $\displaystyle r(x)<Ce^{x}$, for some constant C.
I got everything except I don't know where to get the r(x) part at all...

by letting $\displaystyle u=e^t$, we have:
$\displaystyle f(x)= \int_{x}^{x+1} sin(e^t) \, dt = \int_{e^x}^{e^{x+1}} \frac{\sin u}{u} \, du = e^{y}\int_{e^x}^{e^{x+1}} \sin u \, du $
Thus
$\displaystyle f(x)\leq 2e^{y}\leq 2e^{x}$
where y belongs to the interval (x, x+1).
From the above computation,I think, it should be bounded by a constant C, not Ce^{x}. are you sure it is Ce^{x}? I guess This boundry is too strong .

It does say that in Rudin, if you have it chapter 6 problem 14. But what you did helped. I went about proving the the <2 part a bit incorrectly. Thank you for your help though.

I saw the probelm, maybe you are right, we need a tighter and better boundry.

by using integration by parts twice, you will get the result.