# Intergral

• Dec 10th 2009, 04:34 PM
tedii
Intergral
Let
$f(x)= \int_{x}^{x+1} sin(e^t) \, dt$

Show that
$e^x|f(x)|<2$

and that
$e^x f(x)=cos(e^x)-\frac{cos(e^{x+1})}{e}+r(x)$

where $|r(x)|, for some constant C.

I got everything except I don't know where to get the r(x) part at all...
• Dec 11th 2009, 01:36 AM
Shanks
by letting $u=e^t$, we have:
$f(x)= \int_{x}^{x+1} sin(e^t) \, dt = \int_{e^x}^{e^{x+1}} \frac{\sin u}{u} \, du = e^{-y}\int_{e^x}^{e^{x+1}} \sin u \, du$
Thus
$|f(x)|\leq 2e^{-y}\leq 2e^{-x}$
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 .
• Dec 11th 2009, 11:31 AM
tedii
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.
• Dec 11th 2009, 08:21 PM
Shanks
I saw the probelm, maybe you are right, we need a tighter and better boundry.
• Dec 12th 2009, 01:36 AM
Shanks
by using integration by parts twice, you will get the result.