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| bio | website | qoqosz.net |
|---|---|---|
| location | Poland | |
| age | 24 | |
| visits | member for | 1 year |
| seen | Mar 29 at 12:10 | |
| stats | profile views | 182 |
I'm just a physics student.
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Jun 18 |
revised |
Minimum of integral added 27 characters in body |
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Jun 18 |
answered | Minimum of integral |
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Jun 18 |
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. |
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Jun 18 |
answered | Euler's product formula for $\sin(\pi z)$ and the gamma function |
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Jun 18 |
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 ;). |
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Jun 18 |
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$\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ 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. |
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Jun 18 |
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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. |
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Jun 17 |
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. |
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Jun 17 |
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. |
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Jun 17 |
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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? |
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Jun 17 |
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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 :) |
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Jun 17 |
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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 $2.2641$. |
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Jun 17 |
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The function $f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$ I think it is Ramanujan's formula, i.e.: $$\int\limits_{0}^{\infty}\frac{z^{t} \, dt}{\Gamma(1+t)}=e^{z}-\int\limits_{0}^{\infty}\frac{e^{-z\tau}d\tau}{\tau(\ln^2 (\tau)+\pi^{2})},\,\,\, Re(z)\ge{0}$$ |
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Jun 15 |
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Evaluting: $\int\frac{1}{(1+\tan x)^2} dx$ @experimentX $$\frac{\sin x}{ \sin x + \cos x} - \frac{-\cos x}{\sin x + \cos x} = 1 \neq 2$$ ;) |
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Jun 15 |
answered | Evaluating $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp(n x-\frac{x^2}{2}) \sin(2 \pi x)dx$ |
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Jun 15 |
awarded | Enthusiast |
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Jun 13 |
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Inequality between volume and its projections @nikita2 even obvious facts need a proof in mathematics :) |
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Jun 13 |
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Inequality between volume and its projections @Potato sure it should be measurable. |
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Jun 13 |
awarded | Scholar |
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Jun 13 |
accepted | Inequality between volume and its projections |
