Definite Integrals

[jacks]

Evaluation of definite Integrals

\((a)::\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n}(x)\cdot \sin (nx)dx\)

\((b)::\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\cdot \cos (nx)dx\,,\) where \(n\in \mathbb{N}\)

[Tolaso J Kos]

At least I did not go crazy.. That is a good thing.
Looking over the book "Beyond Series and Integrals" there is a certain technic applied to these kind of integrals. The key is to write the power of \( \cos \) as complex. But that was the first part out of the three tedious parts I had to dwelve.

The second part is to see that the \(\cos \) is even , thus the integral can be re-written with different limits and the final part is using contour integration which was the most tedious thing for me to do. Ok, use contour integration but where? We'll see in a few lines..

Let's begin by re-writting the integral as: \( \displaystyle \int_{0}^{\pi/2}\cos^n x\cos (n x)\,dx=\frac{1}{2}\int_{-\pi/2}^{\pi/2}\cos^n x \cos (n x)\,dx \).
Now the last integral can be re-written as: $$\begin{aligned}
\int_{0}^{\pi/2}\cos^n x \cos(nx)\,dx &=\frac{1}{2}\int_{-\pi/2}^{\pi/2}\cos^n x \cos (nx)\,dx \\
&= \frac{1}{2}\int_{-\pi/2}^{\pi/2}\left ( \frac{e^{ix}+e^{-ix}}{2} \right )^n e^{ix}\,dx\\
&=\frac{1}{2^{n+1}}\int_{-\pi/2}^{\pi/2}\left ( e^{ix}+e^{-ix} \right )^ne^{ix}\,dx \\
&= \frac{1}{2^{n+1}}\int_{-\pi/2}^{\pi/2}\left ( e^{2ix}+1 \right )^n\\
\end{aligned}$$ The obvious sub \(u=2x \) transforms the last integral to \( \displaystyle \frac{1}{2^{n+1}}\int_{-\pi}^{\pi}\left ( e^{iu}+1 \right )^n\,du \).

The latter can be evaluated using contour integration over the unit circle (according to the book).
Subbing \( w=e^{iu} \) we get \( \displaystyle \oint_{\left | w \right |=1}\left ( w+1 \right )^n\frac{dw}{iw}=\frac{\pi}{2^{n+1}} \) by applying Cauchy's Theorem.

Look at this. Who knew that this had closed form? Almost no book refers to this kind of identinties let alone show certain technics on evaluating it.
Jacks , I don't about the first integral but I suppose we will apply the same technic again. I have not dwelved in it further. If someone gets a result I would be more than glad to see his solution.

Phew! Extremely tedious!

Question: Why does the book use contour integration over the unit circle?

[Tolaso J Kos]

An extented summation for the first one. We're distinguishing cases:

\( \bullet \) \( n=\) even . Then it holds: $$\sin^n x= \frac{1}{2^n}\binom{n}{n/2}+\frac{2}{2^n}\sum_{k=0}^{n/2-1}\left ( -1 \right )^{n/2-k}\binom{n}{k}\cos \left [ \left ( n-2k \right )x \right ]$$ Thus our integral yields: $$I = \int_{0}^{\frac{\pi}{2}}\left[\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \cos[(n-2k)x]\right] \cdot \sin(nx) \ dx$$
$$I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \int_{0}^{\frac{\pi}{2}}\sin(nx) \ dx + \frac{2}{2^n} \int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \cos[(n-2k)x] \sin(nx) \ dx$$
$$I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \left[-\frac{\cos(nx)}{n}\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2^n} \int_{0}^{\frac{\pi}{2}}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \left[\sin(2n-2k)x +\sin(2k)x\right] \ dx$$
$$ I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \left[\frac{1}{n}-\frac{\cos(\frac{n\pi}{2})}{n}\right] + \frac{1}{2^n}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \left[-\frac{\cos(2n-2k)x}{2n-2k} -\frac{\cos(2k)x}{2k}\right] _{0}^{\frac{\pi}{2}}$$
$$I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \frac{2\sin^{2}(\frac{n\pi}{4})}{n} + \frac{1}{2^n}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \left[\frac{1}{2n-2k}-\frac{\cos(n-k)\pi}{2n-2k}+\frac{1}{2k} -\frac{\cos(k\pi)}{2k}\right]$$
$$I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \frac{2\sin^{2}(\frac{n\pi}{4})}{n} + \frac{1}{2^n}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \left[\frac{1}{2n-2k}-\frac{\cos(k\pi)}{2n-2k}+\frac{1}{2k} -\frac{\cos(k\pi)}{2k}\right] \;\;\;\; (*) $$
$$I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \frac{2\sin^{2}(\frac{n\pi}{4})}{n} + \frac{1}{2^n}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \left[\frac{2\sin^{2}\frac{k\pi}{2}}{2n-2k}+\frac{2\sin^{2}\frac{k\pi}{2}}{2k}\right]$$
$$ I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \frac{2\sin^{2}(\frac{n\pi}{4})}{n} + \frac{1}{2^n}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \left[\frac{\sin^{2}\frac{k\pi}{2}}{n-k}+\frac{\sin^{2}\frac{k\pi}{2}}{k}\right]$$
$$ I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \frac{2\sin^{2}(\frac{n\pi}{4})}{n} + \frac{1}{2^n}\sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \cdot n \cdot \frac{\sin^{2}\frac{k\pi}{2}}{k(n-k)}$$
$$ I =\frac{1}{2^n} {{n}\choose{\frac{n}{2}}} \frac{2\sin^{2}(\frac{n\pi}{4})}{n} + \frac{1}{2^n}\sum_{k=1}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} {{n}\choose{k}} \cdot n \cdot \frac{\sin^{2}\frac{k\pi}{2}}{k(n-k)}$$

*******************************************************************
\((*)\) Since \( n \) is even, it holds: \( \cos(n-k)\pi = \cos (k\pi) \).

___________________________________________________________________________________
\( \bullet \;\;\; n=\) odd. Then it holds:
$$\sin^n \theta=\frac{2}{2^n}\sum_{k=0}^{(n-1)/2}(-1)^{(n-1)/2-k}\binom{n}{k}\sin \left [ \left ( 2-2k \right )\theta \right ]$$
Unfortunately, I was too lazy to finish up the calculations. Therefore I don't know what it evaluates to.