<s>[INST] how to $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$? 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 $\frac e5 \ln(\pi)< \sqrt{\frac{\pi}{8}}$
 [/INST] The given integral equals
$$\begin{eqnarray*} \int_{0}^{\pi}\sin(x)\sum_{n\geq 0}\frac{(-1)^n}{x+n\pi+1}\,dx &=& \frac{1}{2\pi}\int_{0}^{\pi}\sin(x)\left[\,\psi\left(\tfrac{x+1}{2\pi}+\tfrac{1}{2}\right)-\psi\left(\tfrac{x+1}{2\pi}\right)\right]\,dx\\&=&\int_{\frac{1}{2\pi}}^{\frac{\pi+1}{2\pi}}\sin(2\pi z-1)\left[\,\psi\left(z+\tfrac{1}{2}\right)-\psi(z)\right]\,dz\end{eqnarray*}$$
which can be efficiently approximated by using integration by parts and the Kummer-Malmsten Fourier series:
$$ \log\Gamma(z) = \left(\tfrac12 - z\right)(\gamma + \log 2) + (1 - z)\ln\pi
- \tfrac12\log\sin(\pi z)  + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\sin(2\pi n z)\log n} n$$
holding for any $z\in(0,1)$. Indeed the original integral equals
$$ 2\pi \int_{0}^{1/4}\left[\log\,\frac{\Gamma\left(\tfrac{3}{4}+\tfrac{1}{2\pi}+z\right)\,\Gamma\left(\tfrac{1}{4}+\tfrac{1}{2\pi}-z\right)}{\Gamma\left(\tfrac{3}{4}+\tfrac{1}{2\pi}-z\right)\,\Gamma\left(\tfrac{1}{4}+\tfrac{1}{2\pi}+z\right)}\right]\sin(2\pi z)\,dz $$
where the $\log$ term is a very regular function on the interval $\left(0,\frac{1}{4}\right)$, convex and $$\leq \left(8\log\,\frac{\Gamma\left(\tfrac{1}{2\pi}\right)}{\Gamma\left(\tfrac{1}{2\pi}+\tfrac{1}{2}\right)}-4\log(2\pi)\right)z.$$
</s>