Higher Level Integration Marathon & Questions (1 Viewer)

omegadot · Apr 22, 2017

Re: Extracurricular Integration Marathon

$\noindent If there is still interest in summing a few infinite series you might like to try these two:$

$$\noindent (a) $ \sum^\infty_{n = 1} \frac{1}{n(n + 1)(2n + 1)} \quad $(b) $\sum^\infty_{n = 1} \frac{(-1)^{n + 1}}{n(n + 1)(2n + 1)}.$

seanieg89 · Apr 25, 2017

Re: Extracurricular Integration Marathon

$\frac{1}{n(n+1)(2n+1)}=\frac{1}{n}+\frac{1}{n+1}-\frac{4}{2n+1}\\ \\ \Rightarrow S_1(N)=h_N+h_{N+1}+3-4\sum_{n=0}^N\frac{1}{2n+1}\\ \\=3-4(h_{2N}-h_N)+o(1)\rightarrow 3-4\log(2) $by identification as a Riemann sum.$\\ \\ S_2(N)=-3+4\sum_{n=0}^N \frac{(-1)^{n+1}}{2n+1}+o(1)=\pi-3\\ \\ $(Gregory's series, is very standard and can be proven by integrating the difference of $(1+t^2)^{-1}$ from partial sums of its Maclaurin series and showing this thing tends to zero.)$$

Paradoxica · Apr 27, 2017

Re: Extracurricular Integration Marathon

Evaluate:

$\sum_{k=1}^{\infty} \frac{\left(2\mathcal{H}_{2k} - \mathcal{H}_k\right)^2}{k^2}$

Paradoxica · Apr 27, 2017

Re: Extracurricular Integration Marathon

Hint: Parseval's Theorem

Paradoxica · May 6, 2017

Re: Extracurricular Integration Marathon

Evaluate:

$\int_0^\infty \frac{\log^2{x}}{x^2-x+1} \, \text{d}x$

Hence, or otherwise, evaluate:

$\int_0^\frac{\pi}{3} \log^2{\left( \frac{\sin{\theta}}{\sin{\left(\theta+\frac{\pi}{3}\right)}} \right)} \, \text{d}\theta$

Kingom · May 13, 2017

Re: Extracurricular Integration Marathon

$\begin{align*}a)\int _{ 0 }^{ 1 }{ ln(1-x){ \{ \ln { { (x)\} } } }^{ n-1 }\frac { dx }{ x } }\\b)\int _{ 0 }^{ 1 }{ ln(1+x){ \{ \ln { { (x)\} } } }^{ n-1 }\frac { dx }{ x } }\end{align*}$

Paradoxica · Jun 20, 2017

kawaiipotato · Jun 21, 2017

Paradoxica · Jun 21, 2017

kawaiipotato said:
$\noindent $\int_{0}^{2\pi} \frac{x^2 \sin x }{1-2a \cos x +a^2} \mathrm{d}x$ where $a$ is a constant.$

Nice stack exchange steal.

kawaiipotato · Jun 21, 2017

Paradoxica said:
Nice stack exchange steal.

Haha was redirected there by someone after I couldn't answer it for them

Paradoxica · Oct 14, 2017

kawaiipotato said:
$\noindent $\int_{0}^{2\pi} \frac{x^2 \sin x }{1-2a \cos x +a^2} \mathrm{d}x$ where $a$ is a constant.$

Observe that the term

$\frac{\sin{x}}{a^2+1-2a\cos{x}} \equiv \Im{\left(\frac{1}{a-e^{-ix}}\right)}$

This observation leads us to the immediate conclusion:

Cases must be taken for the sign of log(a²)

For log(a²)<0:

$\int_0^{2\pi} \frac{x^2 \sin{x}}{1-2a\cos{x}+a^2}\text{d}x = \frac{4\pi^2 \log{(1-a)}}{a}$

For log(a²)>0:

$\int_0^{2\pi} \frac{x^2 \sin{x}}{1-2a\cos{x}+a^2}\text{d}x = \frac{4\pi^2 \log{(1-\frac{1}{a})}}{a}$

Proof is left as an exercise to the reader.

Paradoxica · Oct 14, 2017

Re: Extracurricular Integration Marathon

Kingom said:
$\begin{align*}\text{a})\int _{ 0 }^{ 1 }{ \ln(1-x){ \{ \ln { { (x)\} } } }^{ n-1 }\frac { dx }{ x } }\\\text{b})\int _{ 0 }^{ 1 }{ \ln(1+x){ \{ \ln { { (x)\} } } }^{ n-1 }\frac { dx }{ x } }\end{align*}$

By inspection, the answer to the first integral is:

$(-1)^n \zeta{(n+1)} \Gamma{(n)}$

Similarly, the answer to the second integral is:

$(-1)^{n-1} \eta{(n+1)} \Gamma{(n)}$

These come from the integral definitions of the Riemann Zeta and Dirichlet Eta functions, respectively, with slight manipulation.

Paradoxica · Oct 14, 2017

Paradoxica said:
$\int_0^\frac{\pi}{2} \frac{\tan^{-1}{(a\cos^2{x})}}{\cos^2{x}} \, \text{d}x$

The answer to this integral is:

$\frac{\pi a}{\sqrt{2} \sqrt{1+\sqrt{1+a^2}}}$

Some methods to try: DUTIS, Contour Integration

Kingom · Oct 14, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
By inspection, the answer to the first integral is:

$(-1)^n \zeta{(n+1)} \Gamma{(n)}$

Similarly, the answer to the second integral is:

$(-1)^{n-1} \eta{(n+1)} \Gamma{(n)}$

These come from the integral definitions of the Riemann Zeta and Dirichlet Eta functions, respectively, with slight manipulation.

Can you please show working out? I can't really learn from this and what is this "slight manipulation" that you have stated.

Paradoxica · Oct 14, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
Evaluate:

$\sum_{k=1}^{\infty} \frac{\left(2\mathcal{H}_{2k} - \mathcal{H}_k\right)^2}{k^2}$

The function to consider is:

$\log{\left( \tan{\tfrac{x}{4}} \right)} \equiv \frac{2}{\pi}\sum_{k=1}^\infty \frac{(2\mathcal{H}_{2k} - \mathcal{H}_k)\sin{kx}}{k}$

Then by Parseval's Theorem:

$\frac{1}{\pi} \int_{-\pi}^\pi \log^2{\left( \tan{\tfrac{x}{4}} \right)}\text{d}x = \frac{4}{\pi^2} \sum_{k=1}^\infty \left(\frac{2\mathcal{H}_{2k} - \mathcal{H}_k}{k}\right)^2$

The integral evaluates to π³/2, so the sum is π⁴/8

omegadot · Oct 15, 2017

Re: Extracurricular Integration Marathon

Kingom said:
Can you please show working out? I can't really learn from this and what is this "slight manipulation" that you have stated.

$\noindent I will help fill in the details for (a). (b) is done in a similar fashion.$

$\noindent The two important pieces of information needed in order to solve the first integral is the definition of the zeta function $\zeta(s)$ and the gamma function $\Gamma (z)$. These are:$

$\zeta (s) = \sum^\infty_{n = 1} \frac{1}{n^s}$ and $\Gamma (z) = \int^\infty_0 e^{-x} x^{z - 1} \, dx.$

$\noindent We will also make use of the well-known Maclaurin series expansion for the log function $\ln (1 - x)$, which is$

$\ln (1 - x) = - \sum^\infty_{k = 1} \frac{x^k}{k}, \qquad -1 < x \leq 1.$

$\noindent So in our integral, replacing the log term with its Maclaurin series expansion, after interchanging the order of the summation with the integration (this can be done and follows technically from what is known as the Fubini/Tonelli theorems) we have$

$\begin{align*}I &= \int^1_0 \ln (1 - x) [\ln x]^{n - 1} \frac{dx}{x} = -\sum^\infty_{k = 1} \int^1_0 \frac{x^k}{k} [\ln x]^{n - 1} \frac{dx}{x}\end{align*}$

$\noindent Now on making a substitution of $u = -\ln x, du = -dx/x$ and for the limits of integration, as $x \to 0^+, u \to \infty$ and $x = 1, u = 0$ we have$

$\begin{align*} I &= (-1)^n \sum^\infty_{k = 1} \int^\infty_0 e^{-ku} u^{n - 1} \, du.\end{align*}$

$\noindent Next we let $t = ku, dt = k \, du$ while the limits of integration remain unchanged. Thus$

$\begin{align*} I &= (-1)^n \sum^\infty_{k = 1} \frac{1}{k^{n + 1}} \int^\infty_0 e^{-t} t^{n - 1} \, dt = (-1)^n \zeta (n + 1) \Gamma (n),\end{align*}$

$\noindent where the definitions for the zeta and gamma functions given at the beginning have been used.$

$\noindent So after a `slight manipulation' of the integrand we see the result given by Paradoxica can indeed be readily found!$

Paradoxica · Oct 15, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
Evaluate:

$\int_0^\infty \frac{\log^2{x}}{x^2-x+1} \, \text{d}x$

Hence, or otherwise, evaluate:

$\int_0^\frac{\pi}{3} \log^2{\left( \frac{\sin{\theta}}{\sin{\left(\theta+\frac{\pi}{3}\right)}} \right)} \, \text{d}\theta$

anyway let's bump this one, been sitting here for a very long time.

the first one is a trivial exercise in contour integration/special functions manipulation

the second is a matter of a straightforward substitution

omegadot · Oct 15, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
Evaluate:

$\int_0^\infty \frac{\log^2{x}}{x^2-x+1} \, \text{d}x$

$\noindent I will confine myself to the real domain. Not too sure about the `trivial exercise in special functions manipulation' but I guess `trivial' is in the eye of the beholder.$

$\noindent In solving this problem the special functions I am going to make use of is the Hurwitz zeta function $\zeta (s,q)$ which is defined by$

$\zeta (s,z) = \sum^\infty_{n = 0} \frac{1}{(n + z)^s},$

$\noindent the polygamma function $\psi^{(m)}(z)$ which is related to the Hurwitz zeta function by$

$\zeta(1 + m,z) = \frac{(-1)^{m + 1}}{m!} \psi^{(m)}(z),$

$\noindent and the reflection formula for the polygamma function of$

$\psi^{(m)}(1 - z) + (-1)^{n + 1} \psi^{(m)}(z) = (-1)^m \pi \frac{d^m}{dz^m} \cot (\pi z).$

$\noindent So here we go. First note that since the improper integral converges we can write$

$\begin{align*}I &= \int^\infty_0 \frac{\ln^2 x}{x^2 - x + 1} \, dx\\ &= \int^1_0 \frac{\ln^2 x}{x^2 - x + 1} \, dx + \int^\infty_1 \frac{\ln^2 x}{x^2 - x + 1} \, dx\\ &= 2 \int^1_0 \frac{\ln^2 x}{x^2 - x + 1} \, dx,\end{align*}$

$\noindent after letting $x = 1/u$ in the second of the integrals appearing on the right.$

$\noindent Next, noting the factorisation $1 + x^3 = (x + 1)(x^2 - x + 1)$ the integral for $I$, since it converges, can be rewritten as$

$I = 2 \int^1_0 \frac{\ln^2 x}{1 + x^3} \, dx + 2 \int^1_0 \frac{x \ln^2 x}{1 + x^3}$

$\noindent From the well-known result for the geometric series$

$\frac{1}{1 - x} = \sum^\infty_{n = 0} x^n, \qquad |x| < 1,$

$\noindent on replacing $x $ with $-x^3$ we have$

$\frac{1}{1 + x^3} = \sum^\infty_{n = 0} (-1)^n x^{3n}, \qquad |x| < 1.$

$\noindent Replacing this term in each integral with the above sum, after interchanging the summation with the integration we have$

$I = 2 \sum^\infty_{n = 0} (-1)^n \int^1_0 x^{3n} \ln^2 x \, dx + 2 \sum^\infty_{n = 0} (-1)^n \int^1_0 x^{3n+1} \ln^2 x \, dx.$

$\noindent Each of these integrals can be trivially found using integration by parts twice. When done the result is$

$I = \frac{4}{3^3} \sum^\infty_{n = 0} (-1)^n \frac{1}{(n + \frac{1}{3})^3} + \frac{4}{3^3} \sum^\infty_{n = 0} (-1)^n \frac{1}{(n + \frac{2}{3})^3}.$

$\noindent It is in finding closed-formed expressions for these two sums when the fun with special function begins.$

$\noindent As both sums converge absolutely (they are essentially just $p$-series of order three) the terms in each summation can be rearranged without changing the value of its sum.$

$\noindent I will consider the first sum. A similar thing can be done for the second one. Splinting the sum up into two sums, one for when $n$ is even the other for when $n$ is odd. Doing so gives$

$\begin{align*}S_1 &= \sum^\infty_{n = 0} (-1)^n \frac{1}{(n + \frac{1}{3})^3} = \sum^\infty_{n = 0, n \in even} \frac{1}{(n + \frac{1}{3})^3} - \sum^\infty_{n = 0, n \in odd} \frac{1}{(n + \frac{1}{3})^3}\end{align*}$

$\noindent Reindexing, in the first sum let $n \mapsto 2n$ while in the second let $n \mapsto 2n + 1.$ Thus$

$\begin{align*}S_1 &= \sum^\infty_{n = 0} \frac{1}{(2n + \frac{1}{3})^3} - \sum^\infty_{n = 0} \frac{1}{(2n + \frac{4}{3})^3}\\&= \frac{1}{8} \sum^\infty_{n = 0} \frac{1}{(n + \frac{1}{6})^3} - \frac{1}{8} \sum^\infty_{n = 0} \frac{1}{(n + \frac{2}{3})^3}\\&= \frac{1}{8} \left [\zeta(3, \frac{1}{6}) - \zeta (3, \frac{2}{3}) \right ]. \end{align*}$

$\noindent Similarly for the second sum it can be shown that$

$S_2 = \sum^\infty_{n = 0} (-1)^n \frac{1}{(n + \frac{2}{3})^3} = \frac{1}{8} \left [\zeta(3, \frac{1}{3}) - \zeta (3, \frac{5}{6}) \right ].$

$\noindent So for the integral we have$

$\begin{align*}I &= \frac{1}{54} \left [\left (\zeta(3, \frac{1}{6}) - \zeta (3, \frac{2}{3}) \right ) + \left (\zeta(3, \frac{1}{3}) - \zeta (3, \frac{5}{6}) \right )\right ]\end{align*}$

$\noindent By making use of the relation given at the beginning, each of the four Hurwitz zeta functions can be written in terms of the polygamma function as:$

$\begin{align*}\zeta (3,\frac{1}{6}) &= \zeta (1 + 2, \frac{1}{6}) = -\frac{1}{2} \psi^{(2)}(\frac{1}{6})\end{align*}$

$\begin{align*}\zeta (3,\frac{2}{3}) &= \zeta (1 + 2, \frac{2}{3}) = -\frac{1}{2} \psi^{(2)}(\frac{2}{3})\end{align*}$

$\begin{align*}\zeta (3,\frac{1}{3}) &= \zeta (1 + 2, \frac{1}{3}) = -\frac{1}{2} \psi^{(2)}(\frac{1}{3})\end{align*}$

$\begin{align*}\zeta (3,\frac{5}{6}) &= \zeta (1 + 2, \frac{5}{6}) = -\frac{1}{2} \psi^{(2)}(\frac{5}{6})\end{align*}$

$\noindent Thus we can rewrite the integral as$

$\begin{align*}I &= \frac{1}{108} \left [\left (\psi^{(2)}(\frac{2}{3}) - \psi^{(2)}(\frac{1}{3}) \right ) + \left (\psi^{(2)}(\frac{5}{6}) - \psi^{(2)}(\frac{1}{6}) \right ) \right ]\\&= \frac{1}{108} \left [\left (\psi^{(2)}(1 - \frac{1}{3}) - \psi^{(2)}(\frac{1}{3}) \right ) + \left (\psi^{(2)}(1 - \frac{1}{6}) - \psi^{(2)}(\frac{1}{6}) \right ) \right ]\end{align*}$

$\noindent Now applying the reflection formula for the polygamma function we have$

$\psi^{(2)}(1 - \frac{1}{3}) - \psi^{(2)}(\frac{1}{3}) = \pi \frac{d^2}{dz^2} \cot (z\pi) \Big{|}_{z = 1/3} = \frac{8\pi^3}{3 \sqrt{3}}$

$\noindent and$

$\psi^{(2)}(1 - \frac{1}{6}) - \psi^{(2)}(\frac{1}{6}) = \pi \frac{d^2}{dz^2} \cot (z\pi) \Big{|}_{z = 1/6} = 8\pi^3 \sqrt{3}.$

$\noindent So finally we have$

$\begin{align*}I &= \int^\infty_0 \frac{\ln^2 x}{x^2 - x + 1} \, dx = \frac{1}{108} \left [\frac{8 \pi^3}{3 \sqrt{3}} + 8 \pi^3 \sqrt{3} \right ] = \frac{20 \pi^3 \sqrt{3}}{243}.\end{align*}$

Paradoxica · Oct 15, 2017

Re: Extracurricular Integration Marathon

omegadot said:
$\noindent I will confine myself to the real domain. Not too sure about the `trivial exercise in special functions manipulation' but I guess `trivial' is in the eye of the beholder.$

Beta Function manipulation is pretty trivial....

omegadot · Oct 15, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
Beta Function manipulation is pretty trivial....

Beta function, second derivative with respect to its parameter to get rid of the log squared term, beta function reflection formula is then related to the digamma function (and of course its second derivatives will be needed along the way), then a reflection formula for the digamma function and I guess we are home and hosed but is it not nice to see something different

Higher Level Integration Marathon & Questions (1 Viewer)

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