I was going to give this problem to some of my Calc II students, but I realized that I can't even do this without using some tricks involving infinite series (which they do not know yet), and I now want to see if anyone has any ideas about how to do this problem. This is one of the extra problems from a chapter review in a Stewart Calculus book:
Use integration by parts to show that, for all $x > 0$,$$0 < \int_0^\infty \frac{\sin(t)}{\ln(1 + x + t)}dt < \frac{2}{\ln(1 + x)}$$
I've tried most of the obvious choices for $u$, but none of them seem to work out very nicely. I've also tried some algebraic manipulation to turn the integral into
$$-\int\frac{-(1 + x + t)\sin(t)}{(1 + x + t)\ln(1 + x + t)}dt$$
for which you can take $dv = \frac{-1}{(1 + x + t)\ln(1 + x + t)}$ and $u = (1 + x + t)\sin(t)$ to get the nicer equation
$$\int \frac{\sin(t)}{\ln(1 + x + t)}dt = -\left[\frac{(1 + x + t)\sin(t)}{\ln(1 + x + t)} – \int\frac{\sin(t)}{\ln(1 + x + t)}dt – \int\frac{(1 + x + t)\cos(t)}{\ln(1 + x + t)}dt \right]$$
which gives you some cool equations, but is less than helpful. Any input is appreciated. Thank you!
Edit: Than you all for your comments! This problem is from Calculus: Early Transcendentals 6th edition. It should be question 15 on page 523. And the particular challenge that I am having with this problem is finding a way that a Calculus II student might be expected to solve the problem. If we use some more advanced tools from analysis, this problem is not very difficult, but I'm really interested in finding the "lowest level" approach to this. More specifically, an approach that an average Calculus II student might be able to use if they were provided with some problem scaffolding.
Best Answer
I believe this problem can be solved by taking $u = \frac{1}{\log ( 1 + x + t )}$, $dv = \sin (t)$. Then when you do the standard $\int u \,dv = uv - \int v \, du$, the $uv$ will evaluate to be less than the RHS of your original equation, and the $\int v \, du$ can be shown to be positive.