int. by parts proof: 0 < int[0,infty][sin(t)/ln(1+x+t)]dt <

[dts5044]

use integration by parts to prove that 0 < [integral from 0 to infinity] sint * dt / (ln(1 + x + t)) < 2/(ln(1 + x))

the presence of x and t once again confuses me, and I don't really have a clue about this problem. Can anyone help?

 

[stapel]

Since the integral is with respect to "t", the x is, as far as the integral is concerned, a constant. So the proof is "for any fixed x...", so on and so forth.

But there must be some limitations on x, since logs are not defined for non-positive values. For instance, ln(1 + x) is not defined for x < -1. What other information is provided for this exercise?

Thank you!

Eliz.

 

[dts5044]

ahh sorry i didn't notice it said for all x > 0