$0<\int_0^\infty\frac{\sin t}{\ln(1+x+t)} dt<\frac{2}{\ln(1+x)}$ This is my first time posting so please excuse me if I don't follow the proper etiquette. This one is a rather hard problem that was assigned to me for my calculus 2 class. Thank you for your help!
For $x > 0$, prove that:
$$0<\int_0^\infty\frac{\sin t}{\ln(1+x+t)} dt<\frac{2}{\ln(1+x)}$$
 A: $$\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t = \int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + t + 2n\pi)} - \frac{1}{\ln(1 + x + t + (2n + 1)\pi)} \right)\right]$$
Note
$$f_n(t) = \frac{1}{\ln(1 + x + 2n\pi + t)} - \frac{1}{\ln(1 + x + (2n + 1)\pi + t)}.$$
The derivative
$$f_n'(t)  = - \frac{1}{(1 + x + 2n\pi + t)\ln^2(1 + x + 2n\pi + t)} + \frac{1}{(1 + x + (2n + 1)\pi + t)\ln^2(1 + x + (2n + 1)\pi + t)} < 0.$$
Thus $\forall t > 0, \,\, f_n(t) < f_n(0) = \frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}$.
$$\begin{align}\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t &=\int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty f_n(t) \right] \\ 
&< \int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty f_n(0) \right] \\
&= \left(\int_0^\pi \sin(t)\mathrm{d}t \right)\cdot \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}\right) \\
&= 2 \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}\right) 
 \end{align}$$
Consider the partial sum :
$$\begin{align} S_{N} &= \sum_{n=0}^N \left(\frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}\right) \\
&= \frac{1}{\ln(1 + x)} - \sum_{n=0}^{N-1} \left(\frac{1}{\ln(1 + x + (2n + 1)\pi)} - \frac{1}{\ln(1 + x + (2n + 2)\pi)}\right) - \frac{1}{\ln(1 + x + (2N  + 1)\pi)} \\
&<\frac{1}{\ln(1 + x)}. 
\end{align}$$
So 
$$\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t < \frac{2}{\ln(1 + x)}.  $$
