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Evaluate $\int_\pi^\infty \frac{\sin x}{\ln x}$.

I tried to use THE FEYNMAN WAY, but it doesn't help.

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    $\begingroup$ Your integral doesn't converge on the given interval. $\endgroup$ – Dr. Sonnhard Graubner May 28 '18 at 13:57
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    $\begingroup$ math.stackexchange.com/questions/1135362/… integral is convergent $\endgroup$ –  Sintist May 28 '18 at 14:08
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    $\begingroup$ If you're interested in using Feynman, I suspect that introducing a parameter like $f(t) = \int_\pi^\infty \frac{\sin x}{\ln x} e^{-t\ln x} \, dx$ and differentiating to $f'(t) = -\int_\pi^\infty \sin x \, e^{-t\ln x} \, dx$ might be a good start, as this is then simply the integral of $\frac{\sin x}{x^t}$. This is almost certainly simpler, as the integral of $\frac{\sin x}{x}$ is known, but I confess that I do now know how to further simplify this with $x^t$ on the bottom. Perhaps another can help from here. $\endgroup$ – Sputnik May 28 '18 at 14:22
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As @Sputnik commented, introducing the factor $e^{-t\ln x}$ and differentiating with respect to $t$ gives $$\int^\infty_\pi\frac{\sin x}{x^t}dx$$ By the substitution $u=x-\pi$, you get $$-\int^\infty_0(u+\pi)^{-t}\sin(u)du$$

Applying the generalized binomial theorem, the integral equals $$-\sum^{\infty}_{k=0}\frac{(-t)_k}{k!}\pi^{k}\int^\infty_0u^{-t-k}\sin (u)du$$

As mentioned in my previous post, this equals $$-\sum^\infty_{k=0} \frac{(-t)_k}{k!}\pi^{k}\cos(\frac{\pi(t+k)}2)\Gamma (1-t-k)$$

I wonder if one can further simplify it. Moreover, don’t forget to integrate it with respect to $t$ and find the constant!

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