Here is another way: the identity
$$(a+b)^2 - (a-b)^2 = 4ab$$
converts the proposed integral into
$$\frac{1}{4}\left(\int_{-1}^1\,\ln^2(1-x^2)\,dx - \int_{-1}^1\,\ln^2\left(\frac{1+x}{1-x}\right)\,dx\right).$$
Here both integrals are easy to work out.
Search found 33 matches
- Thu Jun 28, 2018 4:47 pm
- Forum: Calculus
- Topic: \(\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\)
- Replies: 5
- Views: 7612
- Wed Jun 20, 2018 2:11 pm
- Forum: Calculus
- Topic: A definite Integral
- Replies: 2
- Views: 4298
Re: A definite Integral
Thanks, Riemann. The Upper limit should be $\pi/2$.
- Fri Jun 15, 2018 7:55 pm
- Forum: Calculus
- Topic: A definite Integral
- Replies: 2
- Views: 4298
A definite Integral
Evaluate
$$\int_0^{\pi/2}\,\frac{x}{\sin x}\,\log(1 - \sin x)\,dx.$$
$$\int_0^{\pi/2}\,\frac{x}{\sin x}\,\log(1 - \sin x)\,dx.$$
- Wed Aug 09, 2017 3:32 pm
- Forum: Calculus
- Topic: A series involving alternating Harmonic numbers
- Replies: 1
- Views: 4815
A series involving alternating Harmonic numbers
For $k \in \mathbb{N}$, let
$$S(k) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+k}\,H_n,$$
where $H_n$ is the $n$-th harmonic number. It is known that
$$S(0) = \frac{\pi^2}{12} - \frac{1}{2}\,\ln^22,\,\,\,\, S(1) = \frac{1}{2}\,\ln^22.$$
Can you find a closed form for $S(k)$ in general?
$$S(k) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+k}\,H_n,$$
where $H_n$ is the $n$-th harmonic number. It is known that
$$S(0) = \frac{\pi^2}{12} - \frac{1}{2}\,\ln^22,\,\,\,\, S(1) = \frac{1}{2}\,\ln^22.$$
Can you find a closed form for $S(k)$ in general?
- Wed Jul 26, 2017 8:05 pm
- Forum: Calculus
- Topic: A series involving Harmonic numbers
- Replies: 2
- Views: 4785
A series involving Harmonic numbers
Show that
$$\sum_{n=1}^\infty\,\frac{(-1)^{n+1}}{(n+1)^2}H_nH_{n+1} = \frac{\pi^4}{480},$$
where $H_n$ is the $n$-th Harmonic number.
$$\sum_{n=1}^\infty\,\frac{(-1)^{n+1}}{(n+1)^2}H_nH_{n+1} = \frac{\pi^4}{480},$$
where $H_n$ is the $n$-th Harmonic number.
- Wed Jul 12, 2017 6:18 pm
- Forum: Calculus
- Topic: An inequality
- Replies: 0
- Views: 3376
An inequality
For $x > 0$, prove that
$$\left(\sum_{n=0}^\infty\frac{1}{(n+x)^2}\right)^2 \geq 2\,\sum_{n=0}^\infty\frac{1}{(n+x)^3}.$$
$$\left(\sum_{n=0}^\infty\frac{1}{(n+x)^2}\right)^2 \geq 2\,\sum_{n=0}^\infty\frac{1}{(n+x)^3}.$$
- Tue Jun 27, 2017 6:26 pm
- Forum: Calculus
- Topic: A generating function involving harmonic number of even index
- Replies: 1
- Views: 3712
Re: A generating function involving harmonic number of even index
Let the proposed series be $f(x)$. Then $$f'(x) = \sum_{n=1}^\infty (-1)^{n-1}H_{2n}x^{2n} = - \sum_{n=1}^\infty H_{2n}(-x^2)^n.$$ Recall that $$\sum_{n=1}^\infty H_nz^n = - \frac{\ln(1-z)}{1-z}.$$ It follows that \begin{eqnarray*} \sum_{n=1}^\infty H_{2n}z^{2n} & = & \frac{1}{2}\,(\sum_{n=1...
- Wed Jun 14, 2017 2:16 am
- Forum: Calculus
- Topic: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
- Replies: 2
- Views: 3527
Re: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
Let the integral be $I_n$. We show that $$I_n = \left\{\begin{array}{ll} \frac{1}{2k} \left(\frac{1}{2k-1} - \frac{1}{2k-2} + \cdots + 1\right), & \mbox{if $n = 2k+1, k \geq 1$}\\ \frac{2\ln 2}{2k-1} + \frac{1}{2k-1}\left(\frac{1}{2k-2} - \frac{1}{2k-3} + \cdots - 1\right), & \mbox{if $n = 2...
- Mon May 22, 2017 4:01 pm
- Forum: Calculus
- Topic: On a series with cosine
- Replies: 2
- Views: 3298
Re: On a series with cosine
Recall that $$\sum_{n=1}^\infty \frac{1}{n}\,x^n = - \ln(1-x).$$ Let $i = \sqrt{-1}$. Then \begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{n 3^n}\,\cos\left(\frac{n\pi}{12}\right) & = & \Re\left\{\sum_{n=1}^\infty \frac{1}{n}\,\left(\frac{e^{i\pi/12}}{3}\right)^n\right\}\\ & = & \Re\lef...
- Tue May 16, 2017 8:45 pm
- Forum: Real Analysis
- Topic: A functional equation
- Replies: 0
- Views: 2177
A functional equation
Find a function $f(x)$ which satisfies the functional equation:
$$f^2(x) = c(x^2 + 1) + f(x^2),$$
where $c$ is a positive constant.
$$f^2(x) = c(x^2 + 1) + f(x^2),$$
where $c$ is a positive constant.