TITLE: how to $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$?
QUESTION [7 upvotes]: I need to prove irreproachably that $$\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$$ . 
With an approximate calculation $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx\approx 0.62145$ and $\frac e5 \ln(\pi)\approx 0.62233$
We can see by Laplace transform that $$\int_{0}^{+\infty}\frac{\sin x}{1+x}\,dx = \int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds $$ and deduce $$\int_{0}^{+\infty}\frac{\sin x}{1+x}\,dx = \int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds\leq  \sqrt{\int_{0}^{+\infty}\frac{ds}{(1+s^2)^2 } } \sqrt{\int_{0}^{+\infty} e^{-2s}  ds} = \sqrt{\frac{\pi}{8}}.$$
But $\frac e5 \ln(\pi)< \sqrt{\frac{\pi}{8}}$

REPLY [4 votes]: Hint :
Use the following identity :
1)
$$\frac{e^{-x}}{{1+x²}}-\frac{e}{5}\frac{(e^{-x}-e^{-\pi x})}{x}<0 $$ for all $x>0$
2)Use the Frullani's integral to find :
$$\int_{0}^{\infty}\frac{e}{5}\frac{(e^{-x}-e^{-\pi x})}{x}=\frac{e}{5}ln(\pi)$$