# Math Help - Prove some identities!

1. 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.

2. 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 ,

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

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

3. 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!

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

5. 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:

Theorem: Let $f:[0,1]\to\mathbb{R}$ be continuous. Then,

$\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^nf(x)\text{ }dx=f(1)$

Proof: It's evident that for each fixed $f\in\mathcal{C}[0,1]$ (in this context, $\mathcal{C}[0,1]$ is continuous functions from $[0,1]\to\mathbb{R}$ with the infinity norm.) the limit $\displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^nf(x)\text{ }dx$ exists. Thus, consider the linear functional

$\displaystyle \varphi:\mathcal{C}[0,1]\to\mathbb{R}:f\mapsto \lim_{n\to\infty}(n+1)\int_0^1 x^n f(x)\text{ }dx$

We first claim that $\varphi$ is continuous (in fact it's Lipschitz). To see this it suffices to note that for any fixed but arbitrary $f_0\in\mathcal{C}[0,1]$ and $\varepsilon>0$ choosing $f\in\mathcal{C}[0,1]$ such that $\|f-f_0\|_{\infty}<\varepsilon$ ensures that

\displaystyle \begin{aligned}\left|\varphi(f)-\varphi(f_0)\right| & =\left|\lim_{n\to\infty}(n+1)\int_0^1x^n (f(x)-f_0(x))\text{ }dx\right|\\ &\leqslant \lim_{n\to\infty}(n+1)\int_0^1 x^{n+}|f(x)-f_0(x)|\text{ }dx\\ &\leqslant \varepsilon\lim_{n\to\infty}(n+1)\int_0^1 x^n\text{ }dx\\ &= \varepsilon\end{aligned}

from where it follows by the arbitrariness of $f_0$ and $\varepsilon$ that $\varphi$ is continuous everywhere on $\mathcal{C}[0,1]$. (note that if someone knows just a litttle functional analysis this would have been clearer since it's fairly easy to verify that $\ker\varphi$ is closed). Note then that if $\displaystyle \sum_{j=0}^{m}\alpha_j x^j$ is a polynomial in $\mathcal{C}[0,1]$ then

\displaystyle \begin{aligned}\varphi(p) &= \lim_{n\to\infty}(n+1)\int_0^1 x^n \sum_{j=0}^{m}\alpha_j x^j\text{ }dx\\ &= \lim_{n\to\infty}(n+1)\int_0^1 \sum_{j=0}^{m}\alpha_j x^{j+n}\text{ }dx\\ &= \lim_{n\to\infty}(n+1)\sum_{j=0}^{m}\frac{\alpha_j }{m+n+1}\\ &= \sum_{j=0}^{m}\alpha_j\\ &= p(1)\end{aligned}

Next recall that the evaluation functional

$\psi:\mathcal{C}[0,1]\to\mathbb{R}:f\mapsto f(1)$

is continuous. Note then that if $\mathbb{P}[0,1]$ is the set of polynomials on $[0,1]$ then

$\displaystyle \left\{f\in\mathcal{C}[0,1]:\varphi(f)=\psi(f)\right\}\supseteq\mathbb{P}[0,1]$

But, we know from the Stone-Weierstrass theorem that $\mathbb{P}[0,1]$ is dense in $\mathcal{C}[0,1]$ from where it follows (from basic topology) that $\psi=\varphi$. The conclusion follows. $\blacksquare$

6. 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 $f'$.

7. 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.

8. 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

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

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

10. 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

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

11. 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 $\displaystyle \ln(\sin(x))=-\ln(2)-\sum_{n=1}^{\infty}\frac{(-1)^n\cos(2nx)}{n}$.

12. 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.

13. 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

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

14. 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.

15. 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 $\displaystyle I(a,b) = 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2-4\left(\tanh^{-1}\left(\frac{b-1}{\sqrt{b^2-1}}\right)\right)^2$.

Page 4 of 8 First 12345678 Last