Search found 59 matches
- Sat Dec 19, 2015 10:38 am
- Forum: Calculus
- Topic: my first post and a cool and challenging integral
- Replies: 3
- Views: 2941
Re: my first post and a cool and challenging integral
Maybe I am missing something simple or direct :? \begin{align*}I&=\int_0^{\infty} \frac{(1-\sin ax)(1-\cos bx)}{x^2}\,dx\\&= \int_0^{\infty} \frac{1-\cos bx - \sin ax + \frac{1}{2}\sin (a+b)x + \frac{1}{2}\sin (a-b)x}{x^2}\,dx\\&= 2\int_0^{\infty} \frac{\sin^2 \frac{bx}{2}}{x^2}\,dx - \i...
- Tue Dec 15, 2015 2:29 pm
- Forum: Calculus
- Topic: An arctan series
- Replies: 2
- Views: 2454
Re: An arctan series
\begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\lef...
- Mon Dec 14, 2015 8:57 pm
- Forum: Linear Algebra
- Topic: Identity with matrix exponential
- Replies: 2
- Views: 2977
Re: Identity with matrix exponential
Hi Papapetros Vaggelis, Nice solution! :D :clap2: Looking at the Eigenvalues indeed is the simplest approach! You derived the characteristic equation to be: $k(k^2+r^2) = 0$, so by Cayley-Hamilton Theorem we have the identity $A^3+r^2A = 0$. Thus in general we have: $$\begin{align*}A^{2n}=(-1)^{n-1}...
- Sat Dec 12, 2015 4:45 pm
- Forum: Calculus
- Topic: Computation of integral
- Replies: 1
- Views: 1912
Re: Computation of integral
We have: $$\begin{align*}\frac{\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha} &= \frac{1}{1+e^{-\pi(x+i\alpha)}}-\frac{1}{1+e^{-\pi(x - i\alpha)}}\\&= 2\sum\limits_{n=1}^{\infty} (-1)^{n}e^{-n\pi x}\sin n\pi\alpha\end{align*}$$ Thus, we have: \begin{align*}\int_{-\infty}^{\infty} \frac{e^{-...
- Fri Dec 11, 2015 9:56 pm
- Forum: Calculus
- Topic: Integral with fractional part
- Replies: 2
- Views: 2487
Re: Integral with fractional part
\begin{align*}\int_1^{\infty} \frac{\{x\}}{x^n}\,dx &= \sum\limits_{k=1}^{\infty}\int_k^{k+1} \frac{\{x\}}{x^n}\,dx\\ &= \sum\limits_{k=1}^{\infty} \int_k^{k+1} \frac{x - k}{x^n}\,dx\\ &= \sum\limits_{k=1}^{\infty} \left[\frac{x^{2-n}}{2-n} - k\frac{x^{1-n}}{1-n}\right]_k^{k+1}\\ &= ...
- Fri Dec 11, 2015 2:15 pm
- Forum: Linear Algebra
- Topic: Identity with matrix exponential
- Replies: 2
- Views: 2977
Identity with matrix exponential
If $a,b,c$ are real numbers (not all $0$) and denote by $\displaystyle r = \sqrt{a^2+b^2+c^2}$, Show that: $$\exp{\left( \begin{matrix} 0 & a & c \\ -a & 0 & b \\ -c & -b & 0 \end{matrix} \right)} = \cos{r \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &...
- Fri Dec 11, 2015 12:35 am
- Forum: Real Analysis
- Topic: An integral with square roots and logarithms
- Replies: 1
- Views: 1956
Re: An integral with square roots and logarithms
We make the change of variable $x \mapsto x^{1/k}$ and denote $\dfrac{n+1}{k} = a$ (say). \begin{align*}&\int_0^1 x^n\log \left(\sqrt{1+x^k} - \sqrt{1-x^k}\right)\,dx\\=& \frac{1}{k}\int_0^1 x^{a-1}\log \left(\sqrt{1+x} - \sqrt{1-x}\right)\,dx\\=& \frac{\log 2}{ak} - \frac{1}{2ak}\int_0^...
- Thu Dec 10, 2015 5:13 pm
- Forum: Calculus
- Topic: $\int_0^\infty \frac{\log x \sin x}{x}\, dx$
- Replies: 2
- Views: 2423
Re: $\int_0^\infty \frac{\log x \sin x}{x}\, dx$
We might begin with the observation that, for $\mathfrak{Re}(s) < 1$, the integral $\displaystyle \int_{\gamma_r} z^{s-1}e^{iz}\,dz$ vanishes over the contour $\gamma_r = [0,r] \cup re^{i[0,\pi/2]}\cup [ir,0]$. Since, \begin{align*}\left|\int_{[re^{i[0,\pi/2]}]} z^{s-1}e^{iz}\,dz\right| &\le \in...
- Thu Dec 10, 2015 2:26 pm
- Forum: Calculus
- Topic: Improper integral with log.
- Replies: 2
- Views: 2438
Re: Improper integral with log.
\begin{align*}\int_0^{\infty} \log^n \left(\frac{e^{x}}{1-e^{x}}\right)\,dx&= (-1)^{n}\int_0^{\infty} \log^n (1-e^{-x})\,dx\\&= (-1)^{n}\int_0^1 \frac{\log^n (1-x)}{x}\,dx \qquad \text{ change of variable } e^{-x} \mapsto x \\&= (-1)^{n}\int_0^1 \frac{\log^n x}{1-x}\,dx = n!\zeta(n+1)\en...