I saw on SE that:  $$\log(\sin x)=-\log(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n} \phantom{a} (0&lt;x&lt;\pi)$$  This is an extremely useful identity, as it helps solve:   $$\int_{0}^{\pi} \log(\sin(x)) dx$$  But how is it derived? From Taylor series, power series? How do I get this?  Even if someone can start me off that would be great.
22552
I saw the two identities  $$ -\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2) $$ and $$ -\log(\cos(x))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}+\log(2) $$ here: .  How can one prove these two identities?
23719
I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to evaluate the underlying indefinite integral.  I considered using the Maclaurin series for the expression $\sin ^{-1}(x)^2$ to evaluate this integral. Using this Maclaurin series, it is easily seen that the problem of evaluating the above integral is equivalent to the problem of evaluating the series $$\sum _{n=1}^{\infty } \frac{2^{2 n-1} H_{n-\frac{1}{2}}}{(1-2 n) n^2 \binom{2 n}{n}}=-0.9393323982...$$ which Mathematica 11 is not able to evaluate directly.  I also considered using the Maclaurin series for the expression $\log \left(1-x^2\right)$ to evaluate the above integral. Using this Macluarin series, it is easily seen that this integral is equal to $$\frac{1}{4} \left(16 \pi  C-\pi ^3-\pi ^2 \log (4)\right)+\sum _{n=1}^{\infty } \frac{\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},n+1;1\right)}{n^2-2 n^3}=-0.939332...,$$ but Mathematica is unable to evaluate the above infinite series.  What is a closed-form evaluation of $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? What techniques can be applied to evaluate this definite integral?
22219
I am asked to prove $e^{(Y \log(1 + Z))} = (1 + Z)^Y$ using the power series definitions for both $\log(Z)$ and $e^Z$.  Our hint is to find a formula for the nth derivative of the power series expansion of $e^{(Y \log(1 + Z))}$ with respect to Z and then use it to find the coefficient on $Z^n$.  I have found that the formula for the nth derivative is $$\frac{(Y)_n}{(1 + Z)^n}\sum_{n = 0}^\infty\frac{1}{n!}\left[Y \log(1 + Z)\right]^n$$ where $(Y)_n$ is the falling factorial.  I am stuck from here though and cannot figure out how to use this to find the coefficient on $Z_n$.  Any help would be greatly appreciated.  Thanks in advance.
99135
Consider $$\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx$$  I thought about making some substitution like : $x^{y}$(it's easy to calculate $$\int_{0}^{1}\frac{x^{b}-x^{a}}{\log(x)}dx = \int_{a}^{b}dy\int_{0}^{1}x^{y}dx=\log(b-1)-\log(a-1)$$).  But I couldn't find substitution like that. Any hints?
48257
Show that: $$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$ $$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$ Thank you in advance. NOTE: The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$, which is a mistake as noted by the comments below.
18688
Problem:  Show that $$\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.$$  My progress:  The problem before this one had me find the Taylor series for $\ln(1-x)$ which was $$-\sum\limits_{n=1}^\infty \frac{x^n}n$$  so I figured I'd use $x=-1$ and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for $x\in(-1, 1)$. And in any case, my calculation isn't going anywhere, since I end up with the series $1-\frac12+\frac13-\frac14\cdots$  Question 1: How can I use this to solve the problem stated initially, and in the title?  Question 2: I can see why $x$ is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?
107390
I stumbled upon this particular expansion that was included in  post.    $$ \displaystyle \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n}$$    This caught my eye because I remember trying to derive a Taylor series for $\arcsin^{2}(x)$ a while ago without much success.  Can anyone prove this or point me to a material that would show a proof of this identity?  EDIT :  Feel free to use any mathematical apparatus at hand. I'm not interested in a proof fit for a certain level, nor am I looking for utmost elegance (though that would be lovely).
94688
I need to find the Fourier series for $$f(x)=\frac{\pi -x}{2}, 0&lt;x&lt;2\pi$$  Since the interval isn't symmetric over $0$, I guess I need to consider $f$'s periodic extension to $\mathbb R$. let's call it $g$. Then $$g(x)=\begin{cases}\frac{\pi -x}{2}, \text{ if } 0&lt;x&lt;\pi\\  \frac{-\pi -x}{2}, \text{ if }{-\pi &lt;x&lt;0}\end{cases}$$  Due to $g$ being odd, the fourier coefficients $a_n$ are all $0$.  And $$b_n=\frac{2}{\pi}\int _0^\pi g(x)\sin(nx)dx=\frac{1}{\pi}\int _0^\pi (x-\pi)\sin (nx)dx$$ According to  $b_n=1/n$ so the Fourier series is $$\sum _{n=1}^{+\infty}\frac{\sin(kx)}{n}$$  But when I check the result of $x=1/2$ I get  which seems to diverge. It also  and a few other $x$'s.  What am I doing wrong?  I just saw . it looks like my answer is correct. am I doing anything wrong in my verification?  EDIT: I realised I forgot to divide by $\pi$. I fixed that but it still doesn't work. I also updated the links.
13679
I don't understand this. These identities are given in the online notes for MIT's 18.01 calculus class. It's related to taking the sum of two trig functions and transforming them into a single trig function. I can use the formulas to do this, but I am having trouble finding anything on the internet to use a proof for my understanding.  Thanks for any responses!  EDITED (added after the first answer):  Sorry, the identity goes like this:  $A\sin{k(x-c)}=a\sin{kx}+b\cos{kx}$ ,  with the relationships  $a=A\cos{kc}$ , $b=-A\sin{kc}$ , $A=\sqrt{a^2+b^2}$ , and $\tan{kc}=-b/a$
98581
I am reading up on the fourier series, and I keep seeing it as being defined as:  $$ f(\theta)= \frac{1}{2}a_0 + \sum_{n=1}^{\infty}(a_n \cos(n\theta) + b_n \sin(n\theta)) $$  where  $$ a_n = \frac{1}{\pi}\int_{0}^{2\pi}\cos(n\theta)f(\theta)d\theta $$  and   $$ b_n = \frac{1}{\pi}\int_{0}^{2\pi}\sin(n\theta)f(\theta)d\theta $$  I understand the derivation of the coefficients using trig integral identities, but I can't find a clear explanation of why $\frac{1}{2}$ is in front of $a_0$. Can anyone help show my why this is the case? Why can't we just have $a_0$ with no number in front of it. Thanks!  edit: corrected summation term
82277
How to prove that   $$ -2\log(2) = -2 + \sum_{n=1}^{\infty}\frac{1}{n(2n+1)} $$  I know that this sum is equal to $\phi(1/2)+\gamma$ where $\phi(x)$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant, but I want to evaluate the sum without knowing it.
57252
How to prove that the series $$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$ is convergent? What about finding the sum?  My attempt: $$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$ The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum.
54750
Show that $$\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)$$ Use this to deduce that the series $$\sum\limits_{k=1}^\infty \frac{1}{k}$$ diverges.   Hint: Use the estimate $$\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x$$   Note that all instances of "$\log$" mean the natural logarithm, the inverse of the exponential. After doing the integration, I've concluded that $\frac{1}{k} \geq \log(1+\frac{1}{k})&gt;-\log(k)$  I suspect I'm supposed to use the hint to arrive at the first inequality, and then since $\log(n)$ can become arbitrarily large as $n$ increases, so must the harmonic series (essentially the comparison test).  I'm having trouble making this jump. Hints are appreciated, but please no solutions.
80834
This is another intriguing formula coming from Ramanujan's letter to G. H. Hardy dated 16th Jan 1913 $$\frac{\log 1}{\sqrt{1}} - \frac{\log 3}{\sqrt{3}} + \frac{\log 5}{\sqrt{5}} - \cdots = \left(\frac{\pi}{4} - \frac{\gamma}{2} + \frac{\log 2\pi}{2}\right)\left(\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}} - \cdots\right)$$ where $$\gamma = \lim_{n \to \infty}\left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} - \log n\right)$$ is the Euler's constant.  This seems really a different kind of series dealing with $\sqrt{n}$ in the denominator and is so unlike the usual series for $\log (1 + x)$ or $\arctan x$ or even Abel's function $\sum x^{n}/n^{2}$. Please help me out in proving the above identity. Any reference having a proof would also be of great help.
68586
Does anybody know how to prove this series?  $$\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$$  I arrived at this through Mathematica.  I tried writing $\log \left(\frac{2n+1}{2n-1}\right)$ as $\int_0^1 \frac{1}{\frac{2n-1}{2}+x}dx$ and $-\sum_{k=1}^\infty \frac{(-2)^k}{k(2n-1)^k}$ but none of them worked.
33857
Suppose $ \alpha, \beta&gt;0 $. Compute:   $$ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $$   Here is what I do: $$\begin{align} \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &amp;= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}\sin (yx)dy\\ &amp;=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}\sin(yx)dx\\ &amp; \\ &amp;   \qquad\text{let $ yx=u $}\\ &amp; \\ &amp;=\int_{\alpha}^{\beta}\frac{1}{y}dy\int_0^{\infty}\sin u du\\ &amp;=\int_{\alpha}^{\beta}\frac{1}{y}dy\left( -\cos u|_{\infty}+\cos u|_0 \right)\\ &amp;=\log\frac{\beta}{\alpha}(-\cos(\infty)+1)\\ &amp;=\log\frac{\beta}{\alpha}-\cos(\infty)\log\frac{\beta}{\alpha} \end{align}$$  But $ \cos(\infty) $ does not exist right? Does it mean the integral actually diverse?  Edit: The question comes from   Who can point out my mistake in the above deduction?
117534
Consider the following series $$S_N = \sum_{k=1}^N \frac{\sin[k\pi/(N+1)]}{k}.$$ What is the asymptotic behavior of this series for $N \to \infty$?  If we replace the sine function with the factor $1$, we get the standard harmonic series which diverges $\sim \ln N$. However, if we replace the sine function with $(-1)^k$, we get a convergent series with the result $\ln 2$. Therefore, I think that $S_N$, defined above, should diverge, but more slowly than the logarithm. Am I correct?