Let f(x) = integral [x to x+1] (sin(e^t)dt).
Show that (e^x) * |f(x)| < 2
and that (e^x) * f(x) = cos (e^x) - (e^-1)cos(e^(x+1)) + r(x) where:
|r(x)| < Ce^-x, C is a constant
integration by parts
The Attempt at a Solution
Well, this integral obviously isn't an elementary function but I don't really care what it's equal to, just that the product of its absolute value and e^x is less than 2 for all x>0.
So I get:
(x+1)sin(e^(x+1)) - xsin(e^x) - integral [x to x+1] (t(e^t)cos(e^t)dt)
<= xsin(e^(x+1)) + sin(e^(x+1)) - xe^x + integral [x to x+1] ((te^t)dt) (since cos is between -1 and 1)
= xsin(e^(x+1)) + sin(e^(x+1)) - xe^x + te^t [evaluated at x to x+1] - integral [x+1 to x] e^t.
= xsin(e^(x+1)) + sin(e^(x+1)) - xe^x + (x+1)e^(x+1) - xe^x - e^(x+1) + e^x
Which doesn't really get me any kind of maximum value on the function if I take the derivative of that function times e^x.