Proof for integral of sin(t)/t+1 dt from x to 0

1. Jan 11, 2009

p3forlife

1. The problem statement, all variables and given/known data
Prove that the integral of sin(t)/t+1 dt from 0 to x is greater than 0 for all x > 0

2. Relevant equations
If f is bounded on [a,b], then f is integrable on [a,b] iff for every epsilon > 0 there exists a partition P of [a,b] s.t. U(f,P) - L(f,P) < epsilon.

3. The attempt at a solution
When you graph sin(t)/t+1 for t>=0, you get a sinusoidal graph with humps that get smaller and smaller, close to the horizontal axis. So using the definition of the integral as the area under the curve, it would make sense that integral of sin(t)/t+1 dt from x to 0 is greater than 0 along the positive x-axis, since the curve becomes almost insignificant as x>0, so the only area that "counts" is the first hump.

I'm stuck on how to do a formal proof, though. Any help would be much appreciated. Moreover, the graph is not bounded. Thanks!

2. Jan 12, 2009

lurflurf

sin(t)/t>-1
so
1+sin(t)/t>0
The integral of a positive function is positive
maybe you mean sin(t)/(t+1)
for this use your idea the integral has local minimums at x=2pi,4pi,6pi,...
if the integral is ever negative a local minimum must be