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Thu Jun 28, 2018 4:47 pm
Forum: Calculus
Topic: $\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx$
Replies: 5
Views: 2512

### Re: $\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx$

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.
Wed Jun 20, 2018 2:11 pm
Forum: Calculus
Topic: A definite Integral
Replies: 2
Views: 1169

### 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: 1169

### A definite Integral

Evaluate
$$\int_0^{\pi/2}\,\frac{x}{\sin x}\,\log(1 - \sin x)\,dx.$$
Wed Aug 09, 2017 3:32 pm
Forum: Calculus
Topic: A series involving alternating Harmonic numbers
Replies: 0
Views: 964

### 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?
Wed Jul 26, 2017 8:05 pm
Forum: Calculus
Topic: A series involving Harmonic numbers
Replies: 2
Views: 1471

### 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.
Wed Jul 12, 2017 6:18 pm
Forum: Calculus
Topic: An inequality
Replies: 0
Views: 938

### An inequality

For $x > 0$, prove that
$$\left(\sum_{n=0}^\infty\frac{1}{(n+x)^2}\right)^2 \geq 2\,\sum_{n=0}^\infty\frac{1}{(n+x)^3}.$$
Tue Jun 27, 2017 6:26 pm
Forum: Calculus
Topic: A generating function involving harmonic number of even index
Replies: 1
Views: 1134

### 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}^\infty...
Wed Jun 14, 2017 2:16 am
Forum: Calculus
Topic: $\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}$
Replies: 2
Views: 1093

Let the integral be $I_n$. We show that $$I_n = \left\{\begin{array}{ll} \frac{1}{2k} \left(\frac{1}{2k-1} - \frac{1}{2k-2} + \cdots + 1\right), & \mbox{if n = 2k+1, k \geq 1}\\ \frac{2\ln 2}{2k-1} + \frac{1}{2k-1}\left(\frac{1}{2k-2} - \frac{1}{2k-3} + \cdots - 1\right), & \mbox{if n = 2k, k \ge... Mon May 22, 2017 4:01 pm Forum: Calculus Topic: On a series with cosine Replies: 2 Views: 1084 ### Re: On a series with cosine Recall that$$\sum_{n=1}^\infty \frac{1}{n}\,x^n = - \ln(1-x).$$Let i = \sqrt{-1}. Then \begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{n 3^n}\,\cos\left(\frac{n\pi}{12}\right) & = & \Re\left\{\sum_{n=1}^\infty \frac{1}{n}\,\left(\frac{e^{i\pi/12}}{3}\right)^n\right\}\\ & = & \Re\left\{-\ln\left(1 -... Tue May 16, 2017 8:45 pm Forum: Real Analysis Topic: A functional equation Replies: 0 Views: 800 ### A functional equation Find a function f(x) which satisfies the functional equation:$$f^2(x) = c(x^2 + 1) + f(x^2),
where $c$ is a positive constant.