2,148 reputation
619
bio website qoqosz.net
location Poland
age 25
visits member for 2 years, 5 months
seen Sep 14 at 15:55

I'm just a physics student.


Jun
23
comment Sum inequality: $\sum_{k=1}^n \frac{\sin k}{k} \le \pi-1$
Possibly related: math.stackexchange.com/questions/13490/…
Jun
22
comment Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$
Practically this solution uses @PeterTamaroff work, without it you could as well write right away $\frac{\pi^2}{8} - \frac{10}{9} \le \frac{1}{8}$ :/
Jun
22
revised Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$
added 209 characters in body
Jun
22
revised Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$
edited body
Jun
22
answered Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$
Jun
22
answered Computing the derivative from the definition
Jun
22
answered Integral as a limit of a sum
Jun
22
answered How to calculate first variation of length of curve?
Jun
22
answered Prove that: $\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$
Jun
21
comment $\int f(x) dx $ is appearing as $\int dx f(x)$. Why?
@AppliedImagination actually I study physics :)
Jun
21
comment $\int f(x) dx $ is appearing as $\int dx f(x)$. Why?
The argument I heard and which is quite convincing is that in physics $f(x)$ can have very long form. So in order not to forget about $dx$ we write it first :)
Jun
21
comment Limit of sum with binomial coefficient
One way is to consider integral: $$\frac{1}{2i} \int_\gamma \sqrt{ \frac{\Gamma (n+1)}{\Gamma (n+1 - z) \Gamma (z+)}} \frac{1}{\sin \pi z} \, dz$$ where $\gamma$ is a rectangle with corners in $-\frac{1}{2} \pm ib$ and $n+\frac{1}{2} \pm ib$ than take limit $n, b \to +\infty$ which requires some manipulations along the road.
Jun
21
comment Aysmptotic relation
@anon sorry, I meant $F$ and $G$.
Jun
21
comment Aysmptotic relation
@anon we can't use de l'Hospital rule if we don't know limits of $f$ and $g$ when $x \to +\infty$.
Jun
20
comment Integration by means of complex analysis
@PeterTamaroff oh, pardon me. Nice job :)
Jun
20
comment Integration by means of complex analysis
@PeterTamaroff you still use P.V. for cosine which simply can't work.
Jun
20
comment Integration by means of complex analysis
@PeterTamaroff I've got some link for you math.stackexchange.com/questions/9402/… btw today I discovered that here is something like FAQ for topics :D
Jun
20
answered Compute: $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$
Jun
20
comment infinite series involving harmonic numbers and zeta
@Chris you mean double summation $\sum_n \sum_m$? I don't think so, because when treating this sum elementary $H_n$ is sum itself.
Jun
19
awarded  Mortarboard