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About the integral $\displaystyle\int_0^{\pi/2}\cos^n{x}\cos{nx} dx$ and $\displaystyle\int_0^{\pi/2} \cos^n{x}\sin{nx} dx$. The value of the first one contains $\pi$, However, I tried to calculate them by integrating by parts as follows.


Let $I(m,n)$ denote the integral $\displaystyle\int_0^{\pi/2} \cos^m{x}\cos{nx} dx$, we have

$$ \begin{split} I(m,n)&=\displaystyle\int_0^{\pi/2} \cos^m{x}\cos{nx}\,dx=\dfrac{1}{n}\displaystyle\int_0^{\pi/2} \cos^m{x}\,d(\sin{nx})\\ &=\dfrac{1}{n}\cos^m{x}\sin{nx}\biggr|_0^{\pi/2}-\dfrac{m}{n}\displaystyle\int_0^{\pi/2} \sin{nx}\cos^{m-1}{x}\sin{x}\,dx\\ &=-\dfrac{m}{n}\displaystyle\int_0^{\pi/2} \sin{nx}\cos^{m-1}{x}\sin{x}\,dx\\ &=\dfrac{m}{n^2}\displaystyle\int_0^{\pi/2} \cos^{m-1}{x}\sin{x}\,d(\cos{nx})\\ &=\dfrac{m}{n^2}\cos^{m-1}{x}\sin{x}\cos^{nx}\biggr|_0^{\pi/2}-\dfrac{m}{n^2}\displaystyle\int_0^{\pi/2} \cos{nx}\,d(\cos^{m-1}{x}\sin{x})\\ &=-\dfrac{m}{n^2}\displaystyle\int_0^{\pi/2}\cos{nx}(\cos^m{x}+(m-1)\cos^{m-2}{x}\sin^2{x})\,dx\\ &=-\dfrac{m}{n^2}\displaystyle\int_0^{\pi/2} \cos{nx}\cos^m{x}\,d{x}-\dfrac{m(m-1)}{n^2}\displaystyle\int_0^{\pi/2} (1-\cos^2{x})\cos^{m-2}{x}\cos{nx}\,d{x}\\ &=\dfrac{m(m-2)}{n^2}\displaystyle\int_0^{\pi/2} \cos{nx}\cos^{m}{x}\,d{x}-\dfrac{m(m-1)}{n^2}\displaystyle\int_0^{\pi/2} \cos^{n-2}{x}\cos{nx}\,d{x}\\ &=\dfrac{m(m-2)}{n^2}I(m,n)-\dfrac{m(m-1)}{n^2}I(m-2,n), \end{split}, $$

That is,

$$I(m,n)=-\dfrac{m(m-1)}{n^2-m^2+2m}I(m-2,n),$$

Then we can calculate $I(m,n)$ by calculating $I(0,n)$ and $I(1,n)$. I can't see where $\pi$ occurs.

Or I'm on the wrong way... ? I don't know. Anyone can help me?

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  • $\begingroup$ your second step seems wrong to me there should be a + sign $\endgroup$ – Deepesh Meena Aug 28 '18 at 21:52
  • $\begingroup$ I think maybe the theory of integration of complex can help you solve this problem. $\endgroup$ – Veritas Julius Aug 28 '18 at 23:36
  • $\begingroup$ Since the answer below only indirectly answers the OP's question, and I am too lazy to write up an answer right now, I summarize as follows: your definition is recursive, depending on knowing the value of at least one integral, namely $I(0,0)=\int_{0}^{\pi/2}dx\,=\frac{\pi}{2}$ $\endgroup$ – Brevan Ellefsen Aug 29 '18 at 9:41
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Maybe nice to have an alternative way to find the integral. Define $A_n, I_n$ and $J_n$ as follows: $$A_n:=\int^{\pi/2}_0 \cos^n(x) e^{inx}\,dx$$ $$I_n:=\int^{\pi/2}_0 \cos^n(x)\cos(nx)\,dx , \ \ \ \ \ \ J_n:=\int^{\pi/2}_0 \cos^n(x)\sin(nx)\,dx $$ This means that $I_n = \Re A_n$ and $J_n=\Im A_n$. So we only have to find $A_n$. Using $\cos(x) = \frac 1 2 \left( e^{ix}+e^{-ix}\right)$ together with the Binomial Theorem one gets \begin{align} A_n &= \int^{\pi/2}_0 \frac 1{2^n}\left( e^{ix}+e^{-ix} \right)^n e^{inx}\,dx \\ &= \int^{\pi/2}_0 \frac 1{2^n} \sum_{k=0}^n \binom{n}{k} e^{i2kx} \,dx\\ &= \frac 1{2^n} \frac{\pi}{2}+ \frac 1{2^n}\sum_{k=1}^n \binom{n}{k} \int^{\pi/2}_0 e^{i2kx}\,dx\\ &=\frac {\pi}{2^{n+1}} + \frac 1{2^n}\sum_{k=1}^n \binom{n}{k} \frac{1}{i2k} \left( e^{ik\pi}-1\right)\\ &=\frac {\pi}{2^{n+1}} + \frac 1{2^n}\sum_{k=1}^n \binom{n}{k} \frac{1}{i2k} \left( (-1)^k-1\right)\\ &=\frac {\pi}{2^{n+1}} +i \frac 1{2^{n}}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} \frac{1}{2k+1} \\ \end{align} Hence: \begin{align} I_n = \frac{\pi}{2^{n+1}}, \ \ \ \ \ J_n = \frac 1{2^{n}}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} \frac{1}{2k+1} \end{align} I don't know if a nice closed form for $J_n$ exists...

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  • $\begingroup$ Interestingly: $$ \frac12\sum_{k=0}^{n-1}\frac{1/2^k}{n-k}=\frac 1{2^{n}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac{\binom{n}{2k+1}}{2k+1} $$ which is not easy to spot. Innovation should be always rewarded (+1). $\endgroup$ – Hazem Orabi Aug 31 '18 at 2:21
  • $\begingroup$ @HazemOrabi Many thanks! Your neat answer served as a "role model" for me. I wasn't comfortable with the final result of mine until I check the equality of the two sums for some values using Mathematica. $\endgroup$ – Shashi Aug 31 '18 at 13:26
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$\underline{\left(1^{\text{st}}\right)\colon}$

$$ \begin{split} I_{(n-1)}&=+\int_{0}^{\pi/2}\cos^{n-1}(x)\,\cos\left((n-1)x\right)\,dx \\ &=+\int_{0}^{\pi/2}\cos^{n-1}(x)\,\left[\sin(x)\,\sin(nx)+\cos(x)\,\cos(nx)\right]\,dx \\ &=+\int_{0}^{\pi/2}\cos^{n-1}(x)\,\sin(x)\,\sin(nx)\,dx\,+\,\int_{0}^{\pi/2}\cos^{n}(x)\,\cos(nx)\,dx \\ &=+\int_{0}^{\pi/2}\cos^{n-1}(x)\,\sin(x)\,\sin(nx)\,dx\,+\,I_{(n)} \\ &\qquad\color{blue}{\left\{u=\sin(nx)\,\Rightarrow\,du=n\,\cos(nx)\,dx\right\}} \\ &\qquad\color{blue}{\left\{dv=\cos^{n-1}(x)\,\sin(x)\,dx\,\Rightarrow\,v=-{\cos^{n}(x)}/{n}\right\}} \\ &=I_{(n)}+\int_{0}^{\pi/2}\cos^{n}(x)\,\cos(nx)\,dx-\frac{1}{n}\cos^{n}(x)\,\sin(nx)\biggr|_{0}^{\pi/2}=2I_{(n)}-0 \\[6mm] I_{(n)}&=\frac{I_{(n-1)}}{2^{1}}=\frac{I_{(n-2)}}{2^{2}}=\,\,\dots\,\,=\frac{I_{(1)}}{2^{n-1}}=\frac{I_{(0)}}{2^{n}} \\ I_{(0)}&=\int_{0}^{\pi/2}dx\,=\frac{\pi}{2}\,\implies\,\color{red}{I_{(n)}=\int_{0}^{\pi/2}\cos^{n}(x)\,\cos(nx)\,dx=\frac{\pi}{2^{n+1}}} \\ \end{split} $$

$\underline{\left(2^{\text{nd}}\right)\colon}$

$$ \begin{split} J_{(n-1)}&=+\int_{0}^{\pi/2}\cos^{n-1}(x)\,\sin\left((n-1)x\right)\,dx \\ &=-\int_{0}^{\pi/2}\cos^{n-1}(x)\,\left[\sin(x)\,\cos(nx)-\cos(x)\,\sin(nx)\right]\,dx \\ &=-\int_{0}^{\pi/2}\cos^{n-1}(x)\,\sin(x)\,\cos(nx)\,dx\,+\,\int_{0}^{\pi/2}\cos^{n}(x)\,\sin(nx)\,dx \\ &=-\int_{0}^{\pi/2}\cos^{n-1}(x)\,\sin(x)\,\cos(nx)\,dx\,+\,J_{(n)} \\ &\qquad\color{blue}{\left\{u=\cos(nx)\,\Rightarrow\,du=-n\,\sin(nx)\,dx\right\}} \\ &\qquad\color{blue}{\left\{dv=\cos^{n-1}(x)\,\sin(x)\,dx\,\Rightarrow\,v=-{\cos^{n}(x)}/{n}\right\}} \\ &=J_{(n)}+\int_{0}^{\pi/2}\cos^{n}(x)\,\cos(nx)\,dx+\frac{1}{n}\cos^{n}(x)\,\cos(nx)\biggr|_{0}^{\pi/2}=2J_{(n)}-\frac{1}{n} \\[6mm] J_{(n)}&=\frac{2^{-1}}{n}+\frac{J_{(n-1)}}{2^{1}}=\frac{2^{-1}}{n}+\frac{2^{-2}}{n-1}+\frac{J_{(n-2)}}{2^{2}}=\,\,\dots\,\,=\sum_{k=0}^{n-1}\frac{2^{-1-k}}{n-k}+\frac{J_{(0)}}{2^{n}} \\ J_{(0)}&=0\,\implies\,\color{red}{J_{(n)}=\int_{0}^{\pi/2}\cos^{n}(x)\,\sin(nx)\,dx=\sum_{k=0}^{n-1}\frac{2^{-1-k}}{n-k}} \\ \end{split} $$ Where the last expression can be written using Lerch Zeta function: $$ J_{(n)} =\frac{1}{2^{n+1}}\sum_{k=0}^{n-1}\frac{2^{n-k}}{n-k} =\frac{1}{2^{n+1}}\sum_{k=1}^{n}\frac{2^k}{k} =-\frac{i\,\pi}{2^{n+1}}-\Phi(2,1,n+1) $$

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