Linked Questions

676
votes
43answers
98k views

Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)

As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler ...
13
votes
4answers
363 views

How can I solve $\int_0^1\frac{\arctan(x^2)}{1+x^2}\,\mathrm dx$?

This integral appears very similar to $\int\frac{\arctan x}{1+x^2}\,\mathrm dx$, but this question cannot be solved through the same simple substitution of $u=\arctan x$. WolframAlpha cannot find a ...
8
votes
8answers
406 views

Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$

I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$ First,I tried to evaluate like this: $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos ...
15
votes
3answers
705 views

Using differentiation under integral sign to calculate a definite integral

I want to calculate the integral $$\int^{\pi/2}_0\frac{\log(1+\sin\phi)}{\sin\phi}d\phi$$ using differentiation with respect to parameter in the integral $$\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\...
5
votes
3answers
470 views

Prove $\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$

$$S=\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$$ Beta function $${1\over {2n\choose n}}=n\int_{0}^{1}x^{n-1}(1-x)^ndx$$ $$\sum_{n=0}^{\infty}{2^n\over {2n\choose ...
14
votes
2answers
581 views

Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $

Evaluate : $$ \int_{0}^{1} \log\left(\dfrac{x^2-2x-4}{x^2+2x-4}\right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} $$ Introduction : I have a friend on another math platform who proposed a summation ...
12
votes
2answers
451 views

Closed-form of $\int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx$

I've conjectured the following closed-form: $$ I = \int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx = -2\,G\,\ln2, $$ where $...
8
votes
2answers
310 views

How can we tackle this integral $\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=-1?$

Something is wrong with this integral (in terms of splitting them out) $$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$ My try: Splitting the ...
17
votes
3answers
364 views

A very odd-looking statement about $\zeta(3)$ and $\text{Li}_2\left(\frac{1}{\varphi^2}\right)$

Some time ago I proved on MSE that the identity $$ \zeta(3)=\frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}\tag{1} $$ (crucial for Apery's proof of $\zeta(3)\not\in\mathbb{Q}$) follows ...
13
votes
1answer
835 views

Closed-form of $\int_0^1 \frac{\ln^2(x)}{\sqrt{x(a-bx)}}\,dx$

I'm interesed in the following integral, for $a,b>0$: $$ \mathcal{I}(a,b) := \int_0^1 \frac{\ln^2(x)}{\sqrt{x(a-bx)}}\,dx $$ Mathematica could evaluate it in term of hypergeometric functions, but I'...
7
votes
3answers
164 views

Prove that $\sum_{n=0}^{\infty}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$

I am seeking alternate proofs for $$\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$$ Here's mine: Recall that, for $x\in(0,2)$, $$\frac1x=\sum_{n\geq0}(1-x)^n$$ Hence we have ...
5
votes
3answers
206 views

Definite integral of $x\sin^n x$ from $0$ to $\pi/2$

How to find \begin{equation*} \int_0^{\pi/2} x\sin^n x dx \end{equation*} where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives \begin{equation*} \frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/...
6
votes
2answers
249 views

A direct proof for $\int_0^x \frac{- x \ln(1-u^2)}{u \sqrt{x^2-u^2}} \, \mathrm{d} u = \arcsin^2(x)$

I have been trying to evaluate $$ f(x) \equiv \int \limits_0^\infty - \ln\left(1 - \frac{x^2}{\cosh^2 (t)}\right) \, \mathrm{d} t $$ for $x \in [0,1]$ and similar integrals recently. I know that $$ \...
5
votes
1answer
718 views

Integral involving Legendre polynomial and $x^n$

I am trying to show that, \begin{align} I = \int_{-1}^1 x^nP_n(x)\,\mathrm{d}x = \frac{2^{n+1}n!n!}{(2n+1)!} \end{align} So far I have done the following. Rodrigues formula is as follows: \begin{...
2
votes
2answers
802 views

Sum of reciprocals of squares - bounding

Recently in class our teacher told us about the evaluating of the sum of reciprocals of squares, that is $\sum_{n=1}^{\infty}\frac{1}{n^2}$. We began with proving that $\sum_{n=1}^{\infty}\frac{1}{n^2}...

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