# Integration of a equation [closed]

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

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

• 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. Jun 3, 2021 at 13:01
• Okay Thanks @AdrianKeister for your help. I will post there my question Jun 3, 2021 at 13:05
• 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/… Jun 3, 2021 at 13:33
• 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. Jun 3, 2021 at 13:36
• Do you mean the variable of integration $a=d_k$ for some $1\le k\le N?$ Jun 3, 2021 at 15:30

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.