Search found 59 matches

by r9m
Wed Mar 09, 2016 8:11 pm
Forum: Real Analysis
Topic: A problem of absolute convergence
Replies: 2
Views: 2917

Re: A problem of absolute convergence

:clap2: :clap2: Cool solution T!
by r9m
Wed Mar 09, 2016 11:00 am
Forum: Calculus
Topic: On a generalization of a classic integral
Replies: 1
Views: 1825

Re: On a generalization of a classic integral

Consider the integral of $\displaystyle f(z) = \frac{\log z}{1+z^n}$ on the contour $\displaystyle \gamma_R := [0,R]\cup Re^{i [0, \frac{2\pi}{n}]} \cup \left[Re^{\frac{2i\pi}{n}},0\right]$, then, as $R \to \infty$, for $n \ge 2$, \begin{align*} 2\pi i \textrm{Res}\left(\frac{\log z}{1+z^n}, z = e^{...
by r9m
Fri Feb 26, 2016 9:00 am
Forum: Calculus
Topic: A Laplace transform
Replies: 2
Views: 2519

Re: A Laplace transform

\begin{align*}I = \int_0^{\infty} e^{-ax}\sin x \sin \left(\sqrt{x}\right)\,dx &= \int_0^{\infty}xe^{-ax^2}\left(\cos \left(x-x^2\right)-\cos \left(x+x^2\right)\right)\,dx\\&=\mathfrak{Re} \int_0^{\infty}x\left(e^{-(a+i)x^2+ix}-e^{-(a+i)x^2-ix}\right)\,dx\\&= \mathfrak{Re} \int_{-\infty}...
by r9m
Fri Feb 26, 2016 8:07 am
Forum: Calculus
Topic: Nested binomial sum
Replies: 1
Views: 1830

Re: Nested binomial sum

We may begin with the beta function identity for non negative integer values of $a,b$, $$\int_0^1 x^{b-a}(1-x)^a\,dx = \frac{1}{(b+1)\binom{b}{a}}$$ Hence, for non-negative integers $a',b'$, \begin{align*}\sum\limits_{a = 0}^{b}\frac{\binom{b}{a}}{\binom{b+b'}{a+a'}} &= (b+b'+1)\sum\limits_{a = ...
by r9m
Wed Feb 24, 2016 4:57 pm
Forum: Calculus
Topic: Alternating Beta sum
Replies: 1
Views: 1762

Re: Alternating Beta sum

\begin{align}\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}B\left(\frac{1}{2},\frac{n}{2}\right) &= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_0^{1} (1-x)^{-1/2} x^{\frac{n}{2} - 1}\,dx\\&= \int_0^{1}\frac{(1-x)^{-1/2} \log \left(1+\sqrt{x}\right)}{x}\,dx\\&= 2\int_0^1 \frac{\l...
by r9m
Sun Feb 14, 2016 11:12 pm
Forum: Calculus
Topic: A tough product
Replies: 3
Views: 3927

Re: A tough product

There'd be a way with Mellin Transform and residue theorem, but I suppose we could start with, \begin{align*}\sum\limits_{n=1}^{N} \log (1+e^{-2\pi n}) &= \sum\limits_{n=1}^{N-1} n\left(\log (1+e^{-2\pi n}) - \log (1+e^{-2\pi (n+1)}) \right) + N\log (1+e^{-2\pi N})\\&= \sum\limits_{n=1}^{N-1...
by r9m
Wed Feb 10, 2016 5:52 pm
Forum: Calculus
Topic: fun-looking log sum. Seen this one before?.
Replies: 1
Views: 1874

Re: fun-looking log sum. Seen this one before?.

Consider the $N$-th partial summation, \begin{align*}S_N&=\sum\limits_{n=1}^{N} \left(2n\log \left(\frac{4n+1}{4n-1}\right) - 1\right) \\&= -N -\frac{1}{2}\sum\limits_{n=1}^{N} \log [(4n+1)(4n-1)] + \frac{1}{2}\sum\limits_{n=1}^{N} (4n+1)\log (4n+1) - (4n-1)\log (4n-1)\\&= -N - \frac{1}{...
by r9m
Mon Feb 08, 2016 12:11 pm
Forum: Real Analysis
Topic: A problem with $\pi(n)$
Replies: 0
Views: 1606

A problem with $\pi(n)$

Are there polynomial functions $P(x), Q(x) \in \mathbb{R}[x]$ that satisfy: $$\int_{0}^{\log n}\frac{P(x)}{Q(x)}\,\mathrm{d}x = \frac{n}{\pi(n)}$$ for infinitely many $n\in \mathbb{N}$, where, $\pi(n)$ is the prime counting function. Something that might be helpful: The Prime Number Theorem: $\displ...
by r9m
Sun Feb 07, 2016 11:16 am
Forum: Calculus
Topic: arcsin integral
Replies: 2
Views: 2336

Re: arcsin integral

\begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \f...
by r9m
Thu Feb 04, 2016 1:15 pm
Forum: Real Analysis
Topic: A limit
Replies: 1
Views: 2062

Re: A limit

Consider a generalization of the problem: For $z \in \mathbb{C}$, and $-1 < \mathfrak{Re}(z) \le 0$, $$\lim\limits_{n \to \infty} \sum\limits_{k=0}^{\infty} \binom{z}{k}e^{-\frac{k}{n}\log k} = 2^z$$ where, $\displaystyle \binom{z}{k} = \frac{1}{k!}\prod\limits_{j=0}^{k-1}(z-j)$ so that in our case ...