2,103 reputation
518
bio website qoqosz.net
location Poland
age 25
visits member for 2 years, 3 months
seen Aug 15 at 21:28

I'm just a physics student.


Jun
19
suggested suggested edit on Question regarding Kuhn-Tucker multiplier
Jun
19
comment Evaluating Integral with Residue Theorem
@Synia sure, if the residues sum up to $0$ why not? Consider for example function $g(z) = \frac{1}{z} - \frac{1}{z-1}$. Then ${\rm Res}_{z=0} g(z) = 1$ and ${\rm Res}_{z=1} g(z) = -1$ so when integrating along adequate contour the integral will be $0$.
Jun
19
answered infinite series involving harmonic numbers and zeta
Jun
19
comment Integration by means of complex analysis
@robjohn right. In the chat room I proposed to have $I'(t) = \int_0^{+\infty} \frac{\cos tx}{1+x^2}\, dx$ than $I(t) = \int_0^{+\infty} \frac{\sin tx}{x(1+x^2)}\,dx$ and ODE is $I''(t) - t^2 I(t) = - \frac{\pi}{2}, \; I(0)=0, I'(0) = \frac{\pi}{2}$ but there is a problem how to solve it in a nice way :)
Jun
19
comment Integration by means of complex analysis
Let's take $$F(a) = \int_0^{+\infty} \frac{\cos (a+x)}{1+x^2} \, dx$$ then we have the desired ODE: $F''(a) + F(a) = 0$ But the problem is to choose initial conditions :| .
Jun
19
revised Evaluating Integral with Residue Theorem
Fixing formating
Jun
19
answered Evaluating Integral with Residue Theorem
Jun
19
suggested suggested edit on Evaluating Integral with Residue Theorem
Jun
19
answered Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
Jun
19
revised How integrals are computed?
Use LaTeX to write equations
Jun
19
suggested suggested edit on How integrals are computed?
Jun
18
comment Integration by means of complex analysis
@PeterTamaroff yes, you should also note that $F''(x)$ exists also only in P.V. sense.
Jun
18
comment Integration by means of complex analysis
@PeterTamaroff I don't follow - what mistake? :)
Jun
18
awarded  Autobiographer
Jun
18
comment Integration by means of complex analysis
Where $$I_1'(u) = - \int_{- \infty}^{+\infty} \frac{\omega \sin (\omega u)}{1 + \omega^2} \, \mbox{d}\omega = - \int_{-\infty}^{+\infty} \frac{t \sin t}{u^2 + t^2} \, \mbox{d}t = - I_1(u)$$ sorry for two comments, but the characters limit... :)
Jun
18
comment Integration by means of complex analysis
@PeterTamaroff your last integral can be reduced to ODE of 1st order by using the following trick: $$I (u) = \int_{-\infty}^{+\infty} \frac{\cos (\omega u)}{1 + \omega^2} \mbox{d}\omega = 2 \int_{0}^{+\infty} \cos (\omega u) \left\{ \int_0^{+\infty} \sin t \, e^{- \omega t} \, \mbox{d}t \right\} \mbox{d}\omega = \\2 \int_0^{+\infty} \int_0^{+\infty} \cos (\omega u) \sin t \, e^{- \omega t} \, \mbox{d}\omega \,\mbox{d}t = 2 \int_0^{+\infty} \frac{t \sin t}{u^2 + t^2} \, \mbox{d}t = \int_{-\infty}^{+\infty} \frac{t \sin t}{u^2 + t^2} \, \mbox{d}t = - I'(u)$$
Jun
18
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.
Jun
18
comment Integration by means of complex analysis
$\frac{\pi}{2}$ cancels each other.
Jun
18
awarded  Analytical
Jun
18
revised Minimum of integral
edited body