$$\int_0^1 \prod_{n\geq 1} (1-x^n) \, dx = \frac{4\pi\sqrt{3}\sinh(\frac{\pi}{3}\sqrt{23})}{\sqrt{23}\cosh(\frac{\pi}{2}\sqrt{23})}$$  This monstrous expression is from Tolaso Network (tolaso.com.gr). I have no idea how to approach it - converting the product to a sum of logarithms does no good, and one cannot switch the order of product/integral either. The product in itself doesn't converge to anything nice either.  I am interested in seeing the proof of the above identity, as well as an explanation of how exactly $\sqrt{23}$ becomes involved in such a deceptive integral. Both real and complex analytic solutions are welcome. A proof without the pentagonal number theorem would be nice as well, since that somewhat trivializes the problem.
100949
Prove $$\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}.$$ In fact, product $(1-x^n)$ is difficult to compute, I hope you can show me some ideas, thank you!
34591
I have found a proof using complex analysis techniques (contour integral, residue theorem, etc.) that shows $$\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$$ for $n\in \mathbb{N}^+\setminus\{1\}$   I wonder if it is possible by using only real analysis to demonstrate this "innocent" result?  Edit A more general result showing that  $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 &lt; a &lt;b$$ can be found in
32440
When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $\sum_{n\geq 0}\binom{2n}{n}^2\frac{1}{16^n(4n+1)}=\frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4 $ is:     $$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx = \tfrac{1}{32}\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^2\tag{A}$$   which might be regarded as a sort of Ahmed's integral under steroids.  I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $\int_{0}^{1}K(x)\,x^{m\pm 1/4}\,dx$, which are associated with peculiar hypergeometric functions.
54998
While trying some problems along with my friends we had difficulty in this question.   True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.   I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that    The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.   This assures that the above product converges. Could anyone help me in finding out where it converges to? And,   Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?
72513
$$\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi\tag1$$  $k\ge0$  Experimental using wolfram integrator, let me to conclude the closed form for $(1)$ is as follow above.  But I don't know how shows it.  How can we prove $(1)$?  We have  $$\prod_{j=0}^{k-1}\cos(2^jx)={\sin(2^kx)\over 2^k\sin(x)}\tag2$$  I am sure how is this product will relate to $(1)$  Setting $k=0$ we have the   $$\int_{0}^{\infty}{\sin(x)\over x}\mathrm dx={\pi\over 2}$$
112091
I'm trying to compute the integral $$ \int_{0}^{1}{x^n}~dx $$ for any given positive integer $n$ using ONLY the Riemann sums. I'd really appreciate it if you could help me out. Thanks.
81392
I want to show that for all $n\ge 2$, it holds that $$ \int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx. $$ I'm not even getting the result for the induction base $(n=2)$: Using integration by parts, I only get $$ \int \cos^2 x\ dx = \cos x \sin x + \int \sin^2 x\ dx. $$ I'm suspecting that I need to use some trigonometric identity here.
108323
In how many ways can we prove the following theorem?     $$I(n):= \int_0^\pi dx \frac{\sin^2(n x)}{\sin^2 x} = n\pi$$   Here $n$ is a nonnegative integer.   The proof I found is by considering $I(n+1)-I(n)$, which can be reduced to $$ g(n):= \int_0^\pi dx \frac{\sin(2 n x) \cos x}{\sin x}. $$ I then proved that $g(n)=g(n+1)$, whence $g(n) = g(1) = \pi$. This completes the proof. I was wondering if there is a more direct way to prove it. By 'direct' I mean without deriving auxiliary recursions.
85942
$$\int_0^1 \frac{\arcsin(x)}{x}dx$$  This is a proposed for a Calculus II exam, and I have absolutely no idea how to solve it. Tried using Frullani or Lobachevsky integrals, or beta and gamma functions, but I can't even find a way to start it. Wolfram Alpha gives a kilometric solution, but I know that cannot be the only answer. Any help appreciated!
111412
Hi I have to show by induction over the number $n$  the following product rule and I have no idea about it. Could someone provide me the proof and a nice explaination:  $$(f\cdot g)^{(n)} (x)= \sum_{k=0}^n\tbinom n k f^{(n-k)} (x) g^{(k)} (x)$$  The $h^{(k)}$ is the $k`th$ derivative of the function $h^{(0)} := h$  I have to use the in my induction proof rhe recursion formula (Pascal Triangle)  $\tbinom n k = \tbinom {n-1} k + \tbinom {n-1} {k-1}$  with $n&gt;k&gt;0$
102811
This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $\prod_{n=1}^\infty \cos\left(\large\frac{x}{2^n}\right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that $$\large\prod_{n=1}^\infty\cos\left(\frac{x}{2^n}\right)=\frac{\sin (x)}{x}.$$  I can't figure out how to prove this.
105433
Just as the product rule can be generalised to the product of more than two functions, i.e. $$\frac{d}{dx} \left [ \prod_{i=1}^k f_i(x) \right ]  = \sum_{i=1}^k \left(\frac{d}{dx} f_i(x) \prod_{j\ne i} f_j(x) \right) = \left(  \prod_{i=1}^k f_i(x) \right) \left( \sum_{i=1}^k \frac{f'_i(x)}{f_i(x)} \right),$$ is there a way to generalise integration by parts to evaluate $$\int f_1(x) f_2(x) \cdots f_n(x) dx \qquad ?$$    For contextual purposes only, I'm trying to evaluate by hand   $$\int x \cosh(x+1)e^x\sin(x) dx \quad  . $$  I know that I could let $u=x \cosh(x+1)$ and $v \ '=e^x\sin(x)$ but that would require integration by parts to be performed at least three times.  Is there a more-efficient way to evaluate this integral using the proposed 'generalisation'?  Thanks!
34259
I think I have found the final answer $\left(4 \left[\frac{2t \cosh(4t) - \sinh(4t)}{t^2}\right]\right)$ but need to verify it.
22319
Using $\text{n}^{\text{th}}$ root of unity  $$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$  Prove that   $$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
24256
Given that $x_n = \displaystyle \prod_{i=1}^n  \frac{2i-1}{2i}$  Then prove that $x_n \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \mathbb Z_+$  What I did was take the logarithm of $x_n$, and I arrived at: $\log{x_n}=\displaystyle \sum_{i=1}^{n} (\log{(2i-1)} - \log{2i}) $  I'd like to know if I proceeded correctly, and thus would like further guidance to solve the problem. However, if I haven't approached the problem correctly, I'd appreciate hints and techniques that are applicable. Please don't post the whole answer because I'd like to work this out on my own. Thanks.
4063
I've been trying to prove   $$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$  I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic series but this just seems surreal. I can't find any logical reasoning behind this identity. I don't understand . Is there any proof that doesn't require the integral transform?
67670
How to evaluate the product:  $$\prod _{n=2}^\infty\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)$$  $$\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)=\dfrac{1}{1-\dfrac{1}{n^2}}$$  How to proceed now?