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 Jun18 comment Integration by means of complex analysis @PeterTamaroff your last integral can be reduced to ODE of 1st order by using the following trick: $$I (u) = \int_{-\infty}^{+\infty} \frac{\cos (\omega u)}{1 + \omega^2} \mbox{d}\omega = 2 \int_{0}^{+\infty} \cos (\omega u) \left\{ \int_0^{+\infty} \sin t \, e^{- \omega t} \, \mbox{d}t \right\} \mbox{d}\omega = \\2 \int_0^{+\infty} \int_0^{+\infty} \cos (\omega u) \sin t \, e^{- \omega t} \, \mbox{d}\omega \,\mbox{d}t = 2 \int_0^{+\infty} \frac{t \sin t}{u^2 + t^2} \, \mbox{d}t = \int_{-\infty}^{+\infty} \frac{t \sin t}{u^2 + t^2} \, \mbox{d}t = - I'(u)$$ Jun18 comment Integration by means of complex analysis Integrand has a removable singularity in $z=0$, so letting $$f(z) = \begin{cases} \frac{1 - \cos z}{z^2 (1+z^2)} & z \neq 0 \\ \frac{1}{2} & z =0 \end{cases}$$ you don't have to consider upper small circle. Jun18 comment Integration by means of complex analysis $\frac{\pi}{2}$ cancels each other. Jun18 awarded Analytical Jun18 revised Minimum of integral edited body Jun18 revised Minimum of integral added 228 characters in body Jun18 comment Minimum of integral @passenger sorry, I edited my post and now bound is a bit less than 12. Equality in C-S holds if $f(x) = A(x+\frac{1}{3})$ but unfortunately there is no $A$ such that $f$ satisfies given conditions. Jun18 revised Minimum of integral added 27 characters in body Jun18 answered Minimum of integral Jun18 comment Why do mathematicians care so much about zeta functions? In this article arxiv.org/abs/1101.3116v1 authors present brief introduction to Riemann Zeta function and Riemann Hypothesis. Moving forward there are several examples of Riemann Hypothesis applications in physics, which of course is closely connected to Zeta function. Jun18 answered Euler's product formula for $\sin(\pi z)$ and the gamma function Jun18 comment $\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral @robjohn $\frac{x}{1+x^2}$ is also bounded but when integrating over $\mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;). Jun18 comment $\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself. Jun18 comment Harmonic oscillator with stochastic forcing Actually for $\psi = \omega$ the solution is $$x(t) = C_1 \sin \omega t + C_2 \omega t - \frac{t \cos \omega t}{2 \omega}$$ so it's well defined. Jun17 comment The function $f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$ @mlbaker I recalled this equation from one of my bookmarks: artofproblemsolving.com/Forum/viewtopic.php?p=1226810#p1226810 . I also found another source which you may find useful: Erdelyi, A., et al., Higher Transcendental Functions, vol 3. You can find it online at apps.nrbook.com/bateman/Vol3.pdf To be more specific look at the page 217 where author introduces your integral and denotes it by $\nu (x)$. It may be a good place to start looking for other references I guess. Jun17 comment Prove the convergence/divergence of $\sum \limits_{k=1}^{\infty} \frac{\tan(k)}{k}$ @Chris yes, I don't understand what you mean by $x$, but your comment looked to me like you were choosing two sequences $(x_k)$ to prove that limit of a function $\tan x / x$ doesn't exist. Jun17 comment Prove the convergence/divergence of $\sum \limits_{k=1}^{\infty} \frac{\tan(k)}{k}$ @Chris aren't you confusing limit of 'function' with limit of a sequence? Jun17 comment The series $\sum\limits_{k=1}^{\infty} \frac1{\sqrt{{k}{(k^2+1)}}}$ @Chris sure, that's very likely. But even though having a good approximation in nice closed form is worth a shot :) Jun17 comment The series $\sum\limits_{k=1}^{\infty} \frac1{\sqrt{{k}{(k^2+1)}}}$ @Chris you can also use integral to estimate the sum: $$\int_{\frac{1}{2}}^{+\infty} \frac{dx}{\sqrt{x(x^2+1)}}$$ Mathematica gives the answer in terms of hypergeometric function: $2 \sqrt{2} \, _2 F_1 \left( \frac{1}{4}, \frac{1}{2} ; \frac{5}{4}, -4 \right) \approx 2.3261$ while the sum is around $2.2641$. Jun17 comment The function $f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$ I think it is Ramanujan's formula, i.e.: $$\int\limits_{0}^{\infty}\frac{z^{t} \, dt}{\Gamma(1+t)}=e^{z}-\int\limits_{0}^{\infty}\frac{e^{-z\tau}d\tau}{\tau(\ln^2 (\tau)+\pi^{2})},\,\,\, Re(z)\ge{0}$$