Problem of the Week #106 - Can you prove this inequality?

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Show that for all $x>0$,

\[0<\int_0^{\infty} \frac{\sin t}{\ln (1+x+t)}\,dt < \frac{2}{\ln(1+x)}.\]

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Hint: [sp]Use integration by parts.[/sp]

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Pranav. You can find his solution below.

[sp]Using Integration by Parts,
$$\begin{aligned}
\int_0^{\infty} \frac{\sin t}{\ln(1+x+t)}\,dt &=\left(\frac{-\cos t}{\ln(1+x+t)}\right|_0^{\infty}-\int_0^{\infty} \frac{\cos t}{\ln^2(1+x+t) (1+x+t)}\,dt\\
&= \frac{1}{\ln(1+x)}-\int_0^{\infty} \frac{\cos t}{\ln^2(1+x+t) (1+x+t)}\,dt\\
\end{aligned}$$
Notice that
$$\int_0^{\infty} \frac{-1}{\ln^2(1+x+t) (1+x+t)}\,dt < \int_0^{\infty} \frac{\cos t}{\ln^2(1+x+t) (1+x+t)}\,dt < \int_0^{\infty} \frac{1}{\ln^2(1+x+t) (1+x+t)}\,dt$$
So our task is to evaluate,
$$\int_0^{\infty} \frac{1}{\ln^2(1+x+t) (1+x+t)}\,dt$$
To evaluate the above, use the substitution $\ln(1+x+t)=u \Rightarrow \dfrac{dt}{1+x+t}=du$, hence,
$$\int_0^{\infty} \frac{1}{\ln^2(1+x+t) (1+x+t)}\,dt=\int_{\ln(1+x)}^{\infty}\frac{du}{u^2}=\frac{1}{\ln(1+x)}$$
Therefore,
$$\frac{-1}{\ln(1+x)}< \int_0^{\infty} \frac{\cos t}{\ln^2(1+x+t) (1+x+t)}\,dt < \frac{1}{\ln(1+x)}$$
$$\Rightarrow 0< \frac{1}{\ln(1+x)}-\int_0^{\infty} \frac{\cos t}{\ln^2(1+x+t) (1+x+t)}\,dt <\frac{2}{\ln(1+x)}$$
$$\Rightarrow 0< \int_0^{\infty} \frac{\sin t}{\ln(1+x+t)}\,dt <\frac{2}{\ln(1+x)}$$
$$\blacksquare$$[/sp]