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

  • 5
    $\begingroup$ Where did you see this inequalty please? $\endgroup$
    – Guy Fsone
    Oct 6 '17 at 14:56
  • 1
    $\begingroup$ 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

Hint : Use the following identity :


$$\frac{e^{-x}}{{1+x²}}-\frac{e}{5}\frac{(e^{-x}-e^{-\pi x})}{x}<0 $$ for all $x>0$

2)Use the Frullani's integral to find :

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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.