Prove some identities!

**Drexel28** · December 26th 2010, 07:55 PM

Originally Posted by chiph588@

How about a little hint on this one..

It's a super contrived question. Consider more generally computing $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n f(x)\text{ }dx$ where $f:[0,1]\to\mathbb{R}$ is continuous.

**simplependulum** · December 26th 2010, 08:29 PM

Originally Posted by Unbeatable0

$\displaystyle \sum_{n=1}^\infty \frac{\text{H}_n}{2^nn}$

Where $\text{H}_n$ is the $n$ 'th harmonic number.

I think of a method that doesn't involve integral calculus but we have to know this fact :

$\displaystyle \frac{1}{1^2} - \frac{1}{2^2} + .... = \frac{\pi^2}{12}$

First note that $\displaystyle H_n = \sum_{k=1}^n \frac{(-1)^{k-1}}{k} \binom{n}{k}$ ( You must know this fact because you have just solved it in this thread ! )

To make it more convenient , i let $\displaystyle \binom{n}{k} = 0$ for $k >n$

Then we have $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{2^n n} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} ~ \frac{\binom{n}{k}}{2^n n }$

$\displaystyle = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{k-1}}{k} ~ \frac{\binom{n}{k}}{2^n n }$

$\displaystyle = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} }{k(k!)} \sum_{n=1}^{\infty} \frac{(n-1)(n-2)...(n-k+1)}{2^n}$

$\displaystyle = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} }{2^k k(k!)} \sum_{n=0}^{\infty} \frac{n(n-1)(n-2)...(n-k+2)}{2^{n-k+1}}$

But we know that $\displaystyle \frac{(k-1)!}{(1-x)^k} = \sum_{n=0}^{\infty} n(n-1)(n-2)...(n-k+2) x^{n-k+1}$ thus ,

$\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{2^n n} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} }{2^k k(k!)} [ 2^k (k-1)! ]$

$\displaystyle = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} }{k^2} = \frac{\pi^2}{12}$

**chiph588@** · December 26th 2010, 08:51 PM

Originally Posted by Drexel28

It's a super contrived question. Consider more generally computing $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n f(x)\text{ }dx$ where $f:[0,1]\to\mathbb{R}$ is continuous.

$\displaystyle \int_0^1 x^n f(x) dx = \frac{f(1)}{n+1}-\frac1{n+1}\int_0^1 x^{n+1} f'(x) dx$ . Since $f$ is continuous over our interval, $f'$ is bounded. Thus $\displaystyle \int_0^1 x^{n+1} f'(x) dx \to 0$ as $n \to \infty$ .

Hence $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n \Gamma^3(x+1)\zeta^3(x+3) dx = \zeta(4)^3$ .

-------

This question was rather deceiving! I was using very complicated formulas for $\Gamma$ and $\zeta$ before the hint.

-------
Random observation: $\displaystyle \zeta(4)^3 \approx \frac4\pi$ . The whole time I was working on this problem I thought the answer was $\frac4\pi$ because of the decimal approximation Wolfram|Alpha gave me. Weird coincidence!

**NonCommAlg** · December 26th 2010, 09:30 PM

Originally Posted by chiph588@

$\displaystyle \int_0^1 x^n f(x) dx = \frac{f(1)}{n+1}-\frac1{n+1}\int_0^1 x^{n+1} f'(x) dx$ . Since $f$ is continuous over our interval, $f'$ is bounded. Thus $\displaystyle \int_0^1 x^{n+1} f'(x) dx \to 0$ as $n \to \infty$ .

Hence $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n \Gamma^3(x+1)\zeta^3(x+3) dx = \zeta(4)^3$ .

$f(1)$ is the answer but your solution is not complete because you've assumed that $f$ is differentiable.

a straightforward solution would be to apply the Weierstrass approximation theorem. it is "straightforward" because obviously for any "polynomial" $p(x)$ we have $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n p(x) dx = p(1).$

**Drexel28** · December 26th 2010, 09:32 PM

Originally Posted by chiph588@

$\displaystyle \int_0^1 x^n f(x) dx = \frac{f(1)}{n+1}-\frac1{n+1}\int_0^1 x^{n+1} f'(x) dx$ . Since $f$ is continuous over our interval, $f'$ is bounded. Thus $\displaystyle \int_0^1 x^{n+1} f'(x) dx \to 0$ as $n \to \infty$ .

Hence $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n \Gamma^3(x+1)\zeta^3(x+3) dx = \zeta(4)^3$ .

-------

This question was rather deceiving! I was using very complicated formulas for $\Gamma$ and $\zeta$ before the hint.

-------
Random observation: $\displaystyle \zeta(4)^3 \approx \frac4\pi$ . The whole time I was working on this problem I thought the answer was $\frac4\pi$ because of the decimal approximation Wolfram|Alpha gave me. Weird coincidence!

This proof doesn't work in general since $f$ is continuous does not imply that $f'$ exists or even if it is differentiable it need not be bounded! Take $f:[0,1]\to\mathbb{R}$ with the obvious continuation of $\displaystyle f(x)=x^2\sin\left(\frac{1}{x^2}\right)$ . It does work though if you consider $f\in C^1[0,1]$ . Try coming up with an alternative.Or, proving that $\Gamma(x+1)\zeta(x+3)\in C^1[0,1]$ . Meanwhile, here is my solution (probably won't be the same as yours, but works out fine):

Spoiler:

**Drexel28** · December 26th 2010, 09:35 PM

Originally Posted by NonCommAlg

$f(1)$ is the answer but your solution is not complete because you've assumed that $f$ is differentiable.

a straightforward solution would be to apply the Weierstrass approximation theorem. it is "straightforward" because obviously for any "polynomial" $p(x)$ we have $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n p(x) dx = p(1).$

Technically his proof does work though since $\Gamma(x+1)\zeta(x+3)\in C^{\infty}[0,1]$ . It just doesn't work for the theorem I stated. Also, as I said in my post even differenetiability doesn't imply boundedness of $f'$ .

**Drexel28** · December 26th 2010, 09:46 PM

Originally Posted by simplependulum

I think of a method that doesn't involve integral calculus but we have to know this fact :

Nice! I thought of doing something like this, but thought it would end up fruitless.

**Drexel28** · December 26th 2010, 10:54 PM

Here are a few challenging ones:

Maybe we should start starring these for level of difficulty? 1 star to 5 scale? Just a thought

Problem $\mathbf{(\star\star\star\star)}$ : Compute

$\displaystyle \int_0^{\frac{\pi}{2}}\ln(\sin(x))\ln(\cos(x))\tex t{ }dx$

Problem $\mathbf{(\star\star)}$ : Compute

$\displaystyle \sum_{k=1}^{\infty}\frac{\sin\left(\frac{k\pi}{3}\ right)}{k^3}$

Problem $\mathbf{(\star\star\star)}$ : Compute

$\displaystyle \int_0^{\frac{\pi}{2}}2\sec(x)\ln\left(\frac{1+\al pha\cos(x)}{1+\beta\cos(x)}\right)\text{ }dx$

**simplependulum** · December 27th 2010, 12:07 AM

Let $\displaystyle I = \int_0^{\pi/2} \ln(\sin(x))\ln(\cos(x))~dx$

$\displaystyle J = \int_0^{\pi/2} \ln^2(\sin(x))~dx$

We can immediately prove that

$\displaystyle J = \int_0^{\pi/2} \ln^2(\cos(x))~dx$

$\displaystyle J = \int_0^{\pi/2} \ln^2(\sin(2x))~dx$ these may be useful in the following .

From $\displaystyle ab = \frac{1}{3} [ (a+b)^2 - \frac{a^2+b^2}{2} - \frac{(a-b)^2}{2} ]$

$\displaystyle = \frac{1}{3}[ (a+b+c)^2 - 2c(a+b) - c^2 - \frac{a^2+b^2}{2} - \frac{(a-b)^2}{2}]$ by letting $c = \ln{2}$ we have

$\displaystyle I = \frac{1}{3} \int_0^{\pi/2} [ \ln^2(\sin(2x)) - 2\ln{2} [\ln(\sin(x)) + \ln(\cos(x)) ]$ $\displaystyle - \ln^2{2} - \frac{\ln^2(\sin(x)) + \ln^2(\cos(x)) }{2} - \frac{ \ln^2(\tan(x)) }{2} ] ~dx$

$\displaystyle = \frac{1}{3} \left[ J +2\pi \ln^2{2} - \frac{\pi \ln^2{2}}{2} - J - \frac{1}{2} \int_0^{\pi/2} \ln^2(\tan(x)) ~dx \right]$

But $\displaystyle \int_0^{\pi/2} \ln^2(\tan(x)) ~dx = \int_0^{\infty} \frac{\ln^2(x)}{1+x^2 }~dx = 4 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^3}$ By using Fourier Series , the value of the summation is $\displaystyle \frac{\pi^3}{32}$

Therefore , $\displaystyle I = \frac{\pi \ln^2{2}}{2} - \frac{\pi^3}{48}$

**Unbeatable0** · December 27th 2010, 05:30 AM

Originally Posted by Drexel28

Problem $\mathbf{(\star\star)}$ : Compute

$\displaystyle \sum_{k=1}^{\infty}\frac{\sin\left(\frac{k\pi}{3}\ right)}{k^3}$

Nice one!

Let

$\displaystyle{ F(x) = \sum_{n=1}^\infty \frac{\sin(nx)}{n^3} }$

on the interval $(0,\pi)$ . Then we have:

$\displaystyle{ F'(x) = \sum_{n=1}^\infty \frac{\cos(nx)}{n^2} }$
$\displaystyle{ F''(x) = -\sum_{n=1}^\infty \frac{\sin(nx)}{n} = \frac{x-\pi}{2} }$

(on the last step I used the familiar fourier series)

Integrating twice and noticing $F(0)=0\:\:\:,\:\:\:F'(0) = \frac{\pi^2}{6}$ , we get:

$\displaystyle{ \sum_{n=1}^\infty \frac{\sin(nx)}{n^3} = F(x) = \frac{1}{12}x^3-\frac{\pi}{4}x^2+\frac{\pi^2}{6}x\:\:\:\:\:\:\:\:\ :\:\:\:\forall x\in[0,\pi) }$

Letting $x=\frac{\pi}{3}$ we get

$\displaystyle{ \sum_{n=1}^\infty \frac{\sin(\frac{\pi n}{3})}{n^3} = \frac{5\pi^3}{162} }$

**Drexel28** · December 27th 2010, 08:39 AM

Good job you two. My solution to the series problem was the same as yours Unbeatable0 but my solution to the one simplependulum was quite different. I made use of the identiy $\displaystyle \ln(\sin(x))=-\ln(2)-\sum_{n=1}^{\infty}\frac{(-1)^n\cos(2nx)}{n}$ .

**Drexel28** · December 27th 2010, 08:51 AM

Originally Posted by Unbeatable0

Nice one!

If you liked that one see how you like

Problem $\mathbf{(\star\star)}$ : Compute

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\pi^2n^2+1}$

and

Problem $\mathbf{(\star\star)}$ : Compute

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2}$

Try different methods for this one! Also, don't forget that other integral that I posted. It's an interesting one.

**Unbeatable0** · December 27th 2010, 12:22 PM

Originally Posted by Drexel28

If you liked that one see how you like

Those sums are known to be easily calculated with the method of residues in complex analysis, and I assume you didn't mean to use this method. I've never seen them calculated in other ways nor do I get an idea of another way from their values. I will, however, keep thinking about doing them in a method not involving complex analysis. It must be interesting to calculate them in such a way.

Originally Posted by Drexel28

Also, don't forget that other integral that I posted. It's an interesting one.

Magic differentiation kills it

We want

$ \displaystyle \int_0^{\frac{\pi}{2}}2\sec(x)\ln\left(\frac{1+\al pha\cos(x)}{1+\beta\cos(x)}\right)dx $

For this, note the integral is $I(\alpha)-I(\beta)$ where

$\displaystyle{ I(\alpha) = \int_0^{\frac{\pi}{2}}2\sec(x)\ln(1+\alpha\cos(x)) dx }$

$\displaystyle{ I'(\alpha) = 2\int_0^{\frac{\pi}{2}}\frac{dx}{1+\alpha\cos{x}} dx = \frac{4\arctan\frac{\alpha-1}{\sqrt{1-\alpha^2}}}{\sqrt{1-\alpha^2}} }$

Calculated by Weierstrass substitution ( $t=\tan\frac{x}{2}$ ).

To calculate the integral of $I'(\alpha)$ put $\alpha = \cos(u)$ and you'll easily get:

$\displaystyle{ I(\alpha)-I(\beta) = \int_\beta^\alpha I'(t)dt = u(u+4\arctan\frac{\cos u-1}{\sin u})\big|^{u=\arccos{\alpha}}_{u=\arccos{\beta}} }$

So our integral equals $J(\alpha)-J(\beta)$ with

$\displaystyle{ J(t) = \arccos(t)\left(\arccos(t)-4\arctan\sqrt\frac{1-t}{1+t}\right) }$

**Drexel28** · December 27th 2010, 12:26 PM

Originally Posted by Unbeatable0

Those sums are known to be easily calculated with the method of residues in complex analysis, and I assume you didn't mean to use this method. I've never seen them calculated in other ways nor do I get an idea of another way from their values. I will, however, keep thinking about doing them in a method not involving complex analysis. It must be interesting to calculate them in such a way.

Yes, the obvious way is to use CA. One question, if you know CA how do you not know multiple integrals?

Magic differentiation kills it

We want

$ \displaystyle \int_0^{\frac{\pi}{2}}2\sec(x)\ln\left(\frac{1+\al pha\cos(x)}{1+\beta\cos(x)}\right)dx $

For this, note the integral is $I(\alpha)-I(\beta)$ where

$\displaystyle{ I(\alpha) = \int_0^{\frac{\pi}{2}}2\sec(x)\ln(1+\alpha\cos(x)) dx }$

$\displaystyle{ I'(\alpha) = 2\int_0^{\frac{\pi}{2}}\frac{dx}{1+\alpha\cos{x}} dx = \frac{4\arctan\frac{\alpha-1}{\sqrt{1-\alpha^2}}}{\sqrt{1-\alpha^2}} }$

Calculated by Weierstrass substitution ( $t=\tan\frac{x}{2}$ ).

To calculate the integral of $I'(\alpha)$ put $\alpha = \cos(u)$ and you'll easily get:

$\displaystyle{ I(\alpha)-I(\beta) = \int_\beta^\alpha I'(t)dt = u(u+4\arctan\frac{\cos u-1}{\sin u})\big|^{u=\arccos{\alpha}}_{u=\arccos{\beta}} }$

So our integral equals $J(\alpha)-J(\beta)$ with

$\displaystyle{ J(t) = \arccos(t)\left(\arccos(t)-4\arctan\sqrt\frac{1-t}{1+t}\right) }$

Woah, your answer looks completely different than mine. Are you sure it's correct? Not doubting your, but it's just really different.

**chiph588@** · December 27th 2010, 12:30 PM

Originally Posted by Drexel28

Woah, your answer looks completely different than mine. Are you sure it's correct? Not doubting your, but it's just really different.

I did magic differentiation too and got $\displaystyle I(a,b) = 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2-4\left(\tanh^{-1}\left(\frac{b-1}{\sqrt{b^2-1}}\right)\right)^2$ .

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