How to prove $$\int_{0}^{\pi}\left|{\sin{nx}\over x}\right| \,\mathrm{d}x\ge{2\over\pi} \sum_{k = 1}^n \frac{1}{k}.$$  I have tried infinite series but absolute value and $n$ ruins it. Induction also didn't work very well.   Any hints?
120611
As my further preparation to Putnam competition, I came across such inequality to prove:  $$\forall n \in \mathbb{N}: \int_{0}^{\pi} \left|\frac{\sin(nx)}{x}\right|dx \geq \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}$$  The problem is that I got stuck in spite of spending by far 6 days on this problem!  Because I got really stuck, I am very determined to see how to prove such inequality. Help very, very appreciated!
68203
Prove, for  every $n\in\mathbb{N}$, that       a) (done) $\displaystyle\int_0 ^{\pi/2} \sin^nx\,dx=\frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}x\,dx$, $n\geq2$.      b) $\displaystyle\int_0^{\pi/2} \sin^{2n+2}x\,dx\leq\int_0^{\pi/2} \sin^{2n+1}x\,dx\leq\int_0^{\pi/2} \sin^{2n}x\,dx.$   So, using part (a)  $$\displaystyle\int_0^{\pi/2} \sin^{2n+2}x\,dx =\frac{2n+1}{2n+2}\int_0^{\pi/2}\sin^{2n}x\,dx\leq\int_0^{\pi/2}\sin^{2n}x\,dx$$  But I don't know how do reduce the expoent in only one unit, since part (a) reduces two units per once.
46932
This refers back to $\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x = \frac\pi2$ already posted. How do I arrive at $\frac\pi2$ using the residue theorem? I'm at the following point: $$\int \frac{e^{iz}}{z} - \int \frac{e^{iz}}{z},$$ and I would appreciate any help.
9047
Prove that if $\phi$ is not equal to $2k\pi$ for any integer $k$, then  $$\sum_{t=0}^{n} \sin{(\theta + t \phi)}=\frac{\sin({\frac{(n+1)\phi}2})\sin{(\theta+\frac{n \phi}2)}}{\sin{(\frac{\phi}2)}}$$  Find a similar formula for  $$\sum_{t=0}^{n}\cos{(\theta+t\phi)}$$  where the functions sin and cos appear on the right-hand side.  Find, for all $\theta$, the values of   $$\sum_{t=0}^{n}\cos^{2}{(2t\theta)}$$ and $$\sum_{t=0}^{n}\sin^{2}{(2t\theta)}$$
115159
Show that for $ab&gt;0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$  I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
114279
Prove that for $ 0&lt;x&lt; \pi $, $$ \quad S_n(x) = \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} &gt;0 \quad \forall n = 1,2,... $$  Having trouble with this problem. This is an olympiad-style question, so an answer that doesn't use calculus or analysis would be preferred. A possible approach is induction, but for this we need to find a function in terms of $n$ and $x$ so that we can actually use the inductive step. If anyone has any ideas they would be appreciated.  If you really want to go down the calculus route (at this point I don't mind), then $ S_n' (x) = \cos x + \cos 3x + ... +\cos((2n-1)x) $ , which you can find a closed form for, but I don't know how useful that is.
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
77661
If $I_n=\int _0^{\pi }\sin^{2n}\theta \:d\theta $, show that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $  Hence calculate $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$  I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.
114161
How to prove that $$ \int_{t_0}^{t_0+T} \sin(m\omega t)\sin(n\omega t)\,\mathrm{d}t$$  will equal to $0$ when $m\ne n$ and $\frac{T}{2}$ when $m=n\ne 0$? Besides $$ \int_{t_0}^{t_0+T} \cos(m\omega t)\cos(n\omega t)\,\mathrm{d}t$$ will equal to $0$ when $m\ne n$ and $\frac{T}{2}$ when $m=n\ne 0$ and $T$ when $m=n=0$?
63713
This  is my problem, true or false?  prove that:   $$\int_{0}^{\infty}\frac{\displaystyle{\sum_{k=1}^{\infty}}k\sin(kx)\,e^{-tk^2}}{\displaystyle{\sum_{k=1}^{\infty}}\cos(kx)\,e^{-tk^2}}dt=\frac{\pi^2({\pi-x})}{8}$$ and $0&lt;x&lt;\dfrac{\pi}{2}$  Thank you.
45228
How to prove   $$ \int_{0}^{\pi/2}\ln\left(\,x^{2} + \ln^{2}\left(\,\cos\left(\,x\,\right)\,\right) \,\right)\,{\rm d}x\ =\ \pi\ln\left(\,\ln\left(\, 2\,\right)\,\right) $$   I don't know how to answer it.  When I asked this integral to my brother, after less than half hours he said it has a nice closed-form involving $\pi$ and $\ln\left(2\right)$ but, as always, he didn't tell me the closed-form and how to obtain it ( I didn't believe him and I think he tried to mess around with me ).  I have also searched the similar question here but it looks like nothing is similar or related.  Could anyone here please help me to obtain the closed form of the integral preferably with elementary ways ( high school methods )?. Any help would be greatly appreciated. Thank you.    Edit:  He is being a little bit nice to me today, he said the closed form is $\pi\ln\ln2$ and it's numerically correct.  This is not a duplicate problem, I am looking for a proof without using complex analysis.
8533
Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$    I think this is known (see ), I appreciate any hint or link for the solution (or the full solution).
50725
Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} &lt; 2 - \frac{1}{n} $  I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
51322
Could please give me a hint.  $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k^{2}+1}$$  It looks like an integral, but I failed to figure out the function.
4558
Show that $$\int\limits_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x =\frac{{{\pi^3}}}{{15}}+2\pi \ln^2 \left({\frac{{1+\sqrt 5 }}{2}}\right).$$   I don't have any idea how to start, but .
1632
A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int\limits_0^\infty \frac{\sin x} x \,\mathrm  dx = \frac \pi 2$$ Well, can anyone prove this without using Residue theory?  I actually thought of using the series representation of $\sin x$: $$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm  dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.
67913
$\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$  Base case: For $n=1$  $sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$  Induction hypothesis: For $n=m$  $\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$  Induction step: $n=m+1$  $\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$  Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$  Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x} {sin\frac{x}{2}}$  How to prove this equality? I used $sin(u)sin(v)$ identity but that didn't help.