Proving bounds of an improper integral

Question

Using integration by parts, show that for all $x > 0$, $$0 < \int_0^\infty \cfrac{\sin{t}}{\ln{(1+x+t)}} \, dt < \frac{2}{\ln{(1+x)}}.$$

Currently, I used integration by parts to get

$$\int_0^\infty \cfrac{\sin{t}}{\ln{(1+x+t)}} = \frac{1}{\ln{(1+x)}} - \int_0^\infty \frac{\cos{t}}{(1+x+t)\ln^2{(1+x+t)}}\, dt.$$

But, I'm not sure how to continue from here...

Answer 1

We have $$ \left |\int\limits_0^\infty {\cos t\over (1+x+t)\ln^2(1+x+t)}\,dt\right |< \int\limits_0^\infty {|\cos t|\over (1+x+t)\ln^2(1+x+t)}\,dt\\ < \int\limits_0^\infty {1\over (1+x+t)\ln^2(1+x+t)}\,dt = -\left.{1\over \ln(1+x+t)}\right\vert_{t=0}^{t=\infty}={1\over \ln(1+x)} $$ This implies both estimates.