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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 $\frac e5 \ln(\pi)< \sqrt{\frac{\pi}{8}}$

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    $\begingroup$ Where did you see this inequalty please? $\endgroup$
    – Guy Fsone
    Oct 6 '17 at 14:56
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    $\begingroup$ Hello gebrane0 there is better I think like this :$$\frac{-y+\phi}{10}+\phi-1$$ where $y$ is the Euler constant and $\phi$ is the golden ratio .A+ $\endgroup$
    – max8128
    Oct 6 '17 at 16:57
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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 :

$$\int_{0}^{\infty}\frac{e}{5}\frac{(e^{-x}-e^{-\pi x})}{x}=\frac{e}{5}ln(\pi)$$

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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 $$\leq \left(8\log\,\frac{\Gamma\left(\tfrac{1}{2\pi}\right)}{\Gamma\left(\tfrac{1}{2\pi}+\tfrac{1}{2}\right)}-4\log(2\pi)\right)z.$$

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  • $\begingroup$ Dear Jack, thank you for these developments, but I do not see how to deduce the inequality $\endgroup$
    – Jane
    Oct 7 '17 at 12:14

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