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$$\int_{x}^{y}\left[\sum_{i=1}^{N}\sqrt{a}\cos\left(\frac{2\pi(d_{i}-a)}{\lambda} \right)\right]^{\!2}da$$

Can anyone solve this integration for me I don't know how the summation and integration will behave with each other

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    $\begingroup$ It's a finite sum as you've written it. You can easily switch the order of integration and summation if you wish. Incidentally, this would be a better question on Mathematics.StackExchange. $\endgroup$ Jun 3, 2021 at 13:01
  • $\begingroup$ Okay Thanks @AdrianKeister for your help. I will post there my question $\endgroup$
    – Parveen
    Jun 3, 2021 at 13:05
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    $\begingroup$ If you do post it elsewhere, it’s best to delete it here. Stack Exchange discourages posting identical questions in multiple places. meta.stackexchange.com/questions/64068/… $\endgroup$ Jun 3, 2021 at 13:33
  • $\begingroup$ Oh, forgot about the fact that the sum is squared. I would probably try to expand out the sum, then see if I could integrate term-by-term. $\endgroup$ Jun 3, 2021 at 13:36
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    $\begingroup$ Do you mean the variable of integration $a=d_k$ for some $1\le k\le N?$ $\endgroup$ Jun 3, 2021 at 15:30

1 Answer 1

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You have to square the sum first. Note that the $\sqrt{a}$ is common to all terms in $$\sqrt{a}\cos\left(\frac{2\pi(d_{i}-a)}{\lambda} \right)\cdot \sqrt{a}\cos\left(\frac{2\pi(d_{j}-a)}{\lambda} \right),$$ so it can factor out as $a.$ That is, we have $$\int_{x}^{y}a\left[\sum_{i=1}^{N}\cos\left(\frac{2\pi(d_{i}-a)}{\lambda} \right)\right]^{\!2}da.$$ We must now square the sum in order to proceed: \begin{align*} &\phantom{=}\left[\sum_{i=1}^{N}\cos\left(\frac{2\pi(d_{i}-a)}{\lambda} \right)\right]^{\!2}\\ &=\sum_{i,j=1}^{N}\cos\left(\frac{2\pi(d_i-a)}{\lambda} \right)\cos\left(\frac{2\pi(d_j-a)}{\lambda} \right)\\ &=\sum_{i,j=1}^{N}\cos\left(\frac{2\pi d_i}{\lambda}-\frac{2\pi a}{\lambda} \right)\cos\left(\frac{2\pi d_j}{\lambda}-\frac{2\pi a}{\lambda}\right)\\ &=\frac12\sum_{i,j=1}^{N}\left[\cos\left(\frac{2\pi d_i}{\lambda}-\frac{2\pi d_j}{\lambda} \right)+\cos\left(\frac{2\pi d_i}{\lambda}+\frac{2\pi d_j}{\lambda}-\frac{4\pi a}{\lambda}\right)\right]\\ &=\frac12\sum_{i,j=1}^{N}\left[\cos\left(\frac{2\pi(d_i-d_j)}{\lambda}\right)+\cos\left(\frac{2\pi(d_i+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)\right]. \end{align*} Now we can move the integral inside the sum: \begin{align*} \int&=\int_x^y\frac{a}{2}\sum_{i,j=1}^{N}\left[\cos\left(\frac{2\pi(d_i-d_j)}{\lambda}\right)+\cos\left(\frac{2\pi(d_i+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)\right]da\\ &=\frac12\sum_{i,j=1}^{N}\int_x^y a\left[\cos\left(\frac{2\pi(d_i-d_j)}{\lambda}\right)+\cos\left(\frac{2\pi(d_i+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)\right]da\\ &=\frac12\sum_{i,j=1}^{N}\left[\int_x^y a\cos\left(\frac{2\pi(d_i-d_j)}{\lambda}\right)da+\int_x^ya\cos\left(\frac{2\pi(d_i+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)da\right]. \end{align*} To continue, we note that according to the comments, there exists $k,\;1\le k\le N$ such that $d_k=a.$ Without loss of generality, we will just assume that $k=N,$ so that $d_N=a.$ Now with the double-sum over $i$ and $j,$ we have four cases to deal with:

  1. $i\not=N, j\not=N.$ The integral is then \begin{align*} &\phantom{=}\frac12\sum_{i,j=1}^{N-1}\left[\int_x^y a\cos\left(\frac{2\pi(d_i-d_j)}{\lambda}\right)da+\int_x^ya\cos\left(\frac{2\pi(d_i+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)da\right]\\ &=\frac12\sum_{i,j=1}^{N-1}\Bigg[\left(\frac{y^2}{2}-\frac{x^2}{2}\right)\cos\left(\frac{2\pi(d_i-d_j)}{\lambda}\right)\\ &\qquad+\int_x^ya\cos\left(\frac{2\pi(d_i+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)da\Bigg] \end{align*}
  2. $i\not=N, j=N.$ For this case, there are $N-1$ terms, and the integral is \begin{align*} &\phantom{=}\frac12\sum_{i=1}^{N-1}\left[\int_x^y a\cos\left(\frac{2\pi(d_i-a)}{\lambda}\right)da+\int_x^ya\cos\left(\frac{2\pi(d_i+a)}{\lambda}-\frac{4\pi a}{\lambda}\right)da\right]\\ &=\frac12\sum_{i=1}^{N-1}\left[\int_x^y a\cos\left(\frac{2\pi(d_i-a)}{\lambda}\right)da+\int_x^ya\cos\left(\frac{2\pi(d_i-a)}{\lambda}\right)da\right]\\ &=\sum_{i=1}^{N-1}\int_x^y a\cos\left(\frac{2\pi(d_i-a)}{\lambda}\right)da. \end{align*}
  3. $i=N, j\not=N.$ For this case, there are $N-1$ terms, and the integral is \begin{align*} &\phantom{=}\frac12\sum_{j=1}^{N-1}\left[\int_x^y a\cos\left(\frac{2\pi(a-d_j)}{\lambda}\right)da+\int_x^ya\cos\left(\frac{2\pi(a+d_j)}{\lambda}-\frac{4\pi a}{\lambda}\right)da\right]\\ &=\frac12\sum_{j=1}^{N-1}\left[\int_x^y a\cos\left(\frac{2\pi(d_j-a)}{\lambda}\right)da+\int_x^ya\cos\left(\frac{2\pi(d_j-a)}{\lambda}\right)da\right]\\ &=\sum_{j=1}^{N-1}\int_x^y a\cos\left(\frac{2\pi(d_j-a)}{\lambda}\right)da. \end{align*} This is the same expression as in Case 2, so we can consolidate these two cases into one case that's doubled: Case 2 and 3: $$2\sum_{j=1}^{N-1}\int_x^y a\cos\left(\frac{2\pi(d_j-a)}{\lambda}\right)da.$$
  4. $i=j=N.$ For this case, there's only $1$ term, and the integral is $$\frac12\left[\int_x^y a\,da+\int_x^ya\,da\right]=\int_x^ya\,da=\frac{y^2}{2}-\frac{x^2}{2}.$$

This is getting rather unwieldy to continue writing down. I would just remark that the integral $$\int_x^y a\cos(c+a)\,da=\cos(c+y)+y\sin(c+y)-\cos(c+x)-x\sin(c+x).$$ All the remaining integrals are of this form. I'll let you finish.

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