I think of a method that doesn't involve integral calculus but we have to know this fact :
$\displaystyle \displaystyle \frac{1}{1^2} - \frac{1}{2^2} + .... = \frac{\pi^2}{12} $
First note that $\displaystyle \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 \displaystyle \binom{n}{k} = 0 $ for $\displaystyle k >n $
Then we have $\displaystyle \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 \displaystyle = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{k-1}}{k} ~ \frac{\binom{n}{k}}{2^n n } $
$\displaystyle \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 \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 \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 \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 \displaystyle = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} }{k^2} = \frac{\pi^2}{12} $
$\displaystyle \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 $\displaystyle f $ is continuous over our interval, $\displaystyle f' $ is bounded. Thus $\displaystyle \displaystyle \int_0^1 x^{n+1} f'(x) dx \to 0 $ as $\displaystyle n \to \infty $.
Hence $\displaystyle \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 $\displaystyle \Gamma $ and $\displaystyle \zeta $ before the hint.
-------
Random observation: $\displaystyle \displaystyle \zeta(4)^3 \approx \frac4\pi $. The whole time I was working on this problem I thought the answer was $\displaystyle \frac4\pi $ because of the decimal approximation Wolfram|Alpha gave me. Weird coincidence!
$\displaystyle f(1)$ is the answer but your solution is not complete because you've assumed that $\displaystyle f$ is differentiable.
a straightforward solution would be to apply the Weierstrass approximation theorem. it is "straightforward" because obviously for any "polynomial" $\displaystyle p(x)$ we have $\displaystyle \displaystyle \lim_{n\to\infty}(n+1)\int_0^1 x^n p(x) dx = p(1).$
This proof doesn't work in general since $\displaystyle f$ is continuous does not imply that $\displaystyle f'$ exists or even if it is differentiable it need not be bounded! Take $\displaystyle f:[0,1]\to\mathbb{R}$ with the obvious continuation of $\displaystyle \displaystyle f(x)=x^2\sin\left(\frac{1}{x^2}\right)$. It does work though if you consider $\displaystyle f\in C^1[0,1]$. Try coming up with an alternative.Or, proving that $\displaystyle \Gamma(x+1)\zeta(x+3)\in C^1[0,1]$. Meanwhile, here is my solution (probably won't be the same as yours, but works out fine):
Spoiler:
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$\displaystyle \mathbf{(\star\star\star\star)}$: Compute
$\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}\ln(\sin(x))\ln(\cos(x))\tex t{ }dx$
Problem$\displaystyle \mathbf{(\star\star)}$: Compute
$\displaystyle \displaystyle \sum_{k=1}^{\infty}\frac{\sin\left(\frac{k\pi}{3}\ right)}{k^3}$
Problem$\displaystyle \mathbf{(\star\star\star)}$: Compute
$\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}2\sec(x)\ln\left(\frac{1+\al pha\cos(x)}{1+\beta\cos(x)}\right)\text{ }dx$
Let $\displaystyle \displaystyle I = \int_0^{\pi/2} \ln(\sin(x))\ln(\cos(x))~dx $
$\displaystyle \displaystyle J = \int_0^{\pi/2} \ln^2(\sin(x))~dx $
We can immediately prove that
$\displaystyle \displaystyle J = \int_0^{\pi/2} \ln^2(\cos(x))~dx $
$\displaystyle \displaystyle J = \int_0^{\pi/2} \ln^2(\sin(2x))~dx $ these may be useful in the following .
From $\displaystyle \displaystyle ab = \frac{1}{3} [ (a+b)^2 - \frac{a^2+b^2}{2} - \frac{(a-b)^2}{2} ] $
$\displaystyle \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 $\displaystyle c = \ln{2} $ we have
$\displaystyle \displaystyle I = \frac{1}{3} \int_0^{\pi/2} [ \ln^2(\sin(2x)) - 2\ln{2} [\ln(\sin(x)) + \ln(\cos(x)) ] $ $\displaystyle \displaystyle - \ln^2{2} - \frac{\ln^2(\sin(x)) + \ln^2(\cos(x)) }{2} - \frac{ \ln^2(\tan(x)) }{2} ] ~dx $
$\displaystyle \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 \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 $\displaystyle \displaystyle \frac{\pi^3}{32}$
Therefore , $\displaystyle \displaystyle I = \frac{\pi \ln^2{2}}{2} - \frac{\pi^3}{48} $
Nice one!
Let
$\displaystyle \displaystyle{
F(x) = \sum_{n=1}^\infty \frac{\sin(nx)}{n^3}
}$
on the interval $\displaystyle (0,\pi)$. Then we have:
$\displaystyle \displaystyle{
F'(x) = \sum_{n=1}^\infty \frac{\cos(nx)}{n^2}
}$
$\displaystyle \displaystyle{
F''(x) = -\sum_{n=1}^\infty \frac{\sin(nx)}{n} = \frac{x-\pi}{2}
}$
(on the last step I used the familiar fourier series)
Integrating twice and noticing $\displaystyle F(0)=0\:\:\:,\:\:\:F'(0) = \frac{\pi^2}{6}$, we get:
$\displaystyle \displaystyle{
\sum_{n=1}^\infty \frac{\sin(nx)}{n^3} = F(x) = \frac{1}{12}x^3-\frac{\pi}{4}x^2+\frac{\pi^2}{6}x\:\:\:\:\:\:\:\:\ :\:\:\:\forall x\in[0,\pi)
}$
Letting $\displaystyle x=\frac{\pi}{3}$ we get
$\displaystyle \displaystyle{
\sum_{n=1}^\infty \frac{\sin(\frac{\pi n}{3})}{n^3} = \frac{5\pi^3}{162}
}$
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 $\displaystyle \displaystyle \ln(\sin(x))=-\ln(2)-\sum_{n=1}^{\infty}\frac{(-1)^n\cos(2nx)}{n}$.
If you liked that one see how you like
Problem$\displaystyle \mathbf{(\star\star)}$: Compute
$\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\pi^2n^2+1}$
and
Problem$\displaystyle \mathbf{(\star\star)}$: Compute
$\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2}$
Try different methods for this one! Also, don't forget that other integral that I posted. It's an interesting one.
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.
Magic differentiation kills it
We want
$\displaystyle
\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 $\displaystyle I(\alpha)-I(\beta)$ where
$\displaystyle \displaystyle{
I(\alpha) = \int_0^{\frac{\pi}{2}}2\sec(x)\ln(1+\alpha\cos(x)) dx
}$
$\displaystyle \displaystyle{
I'(\alpha) = 2\int_0^{\frac{\pi}{2}}\frac{dx}{1+\alpha\cos{x}} dx = \frac{4\arctan\frac{\alpha-1}{\sqrt{1-\alpha^2}}}{\sqrt{1-\alpha^2}}
}$
Calculated by Weierstrass substitution ($\displaystyle t=\tan\frac{x}{2}$).
To calculate the integral of $\displaystyle I'(\alpha)$ put $\displaystyle \alpha = \cos(u)$ and you'll easily get:
$\displaystyle \displaystyle{
I(\alpha)-I(\beta) = \int_\beta^\alpha I'(t)dt = u(u+4\arctan\frac{\cos u-1}{\sin u})\big|^{u=\arccos{\alpha}}_{u=\arccos{\beta}}
}$
So our integral equals $\displaystyle J(\alpha)-J(\beta)$ with
$\displaystyle \displaystyle{
J(t) = \arccos(t)\left(\arccos(t)-4\arctan\sqrt\frac{1-t}{1+t}\right)
}$
Yes, the obvious way is to use CA. One question, if you know CA how do you not know multiple integrals?
Woah, your answer looks completely different than mine. Are you sure it's correct? Not doubting your, but it's just really different.
Magic differentiation kills it
We want
$\displaystyle
\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 $\displaystyle I(\alpha)-I(\beta)$ where
$\displaystyle \displaystyle{
I(\alpha) = \int_0^{\frac{\pi}{2}}2\sec(x)\ln(1+\alpha\cos(x)) dx
}$
$\displaystyle \displaystyle{
I'(\alpha) = 2\int_0^{\frac{\pi}{2}}\frac{dx}{1+\alpha\cos{x}} dx = \frac{4\arctan\frac{\alpha-1}{\sqrt{1-\alpha^2}}}{\sqrt{1-\alpha^2}}
}$
Calculated by Weierstrass substitution ($\displaystyle t=\tan\frac{x}{2}$).
To calculate the integral of $\displaystyle I'(\alpha)$ put $\displaystyle \alpha = \cos(u)$ and you'll easily get:
$\displaystyle \displaystyle{
I(\alpha)-I(\beta) = \int_\beta^\alpha I'(t)dt = u(u+4\arctan\frac{\cos u-1}{\sin u})\big|^{u=\arccos{\alpha}}_{u=\arccos{\beta}}
}$
So our integral equals $\displaystyle J(\alpha)-J(\beta)$ with
$\displaystyle \displaystyle{
J(t) = \arccos(t)\left(\arccos(t)-4\arctan\sqrt\frac{1-t}{1+t}\right)
}$