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How to solve this integration? $\int_0^\pi (\sin x +2\sin^2 x+3\sin^3 x+\dots +100\sin^{100} x)^2 dx$

I tried to solve by using geometric sequence but couldn't solve it. Does anyone have an idea?

Selena
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  • Welcome to MSE. Here is an idea which might help. Notice that inside the brackets, you can factor out $\sin x$, then divide by $\cos x$ on the outside and multiply by $\cos x$ on the all of the terms on the inside to get $\frac{\sin x}{\cos x}\left(\cos x + 2 \cos x \sin x + 3 \cos x \sin^2 x + \cdots 100 \cos x \sin^{99} x \right)$. Next, notice the factor outside is $\tan x$ and each term inside is the derivative of a power of $\sin x$. However, based on the other comment & the answer, their suggestions will likely be easier to use. – John Omielan Dec 27 '18 at 05:28
  • Do you know the sum of series $\sum_{k=1}^{n}kx^{k-1}$? – Paramanand Singh Dec 27 '18 at 05:28
  • @JohnOmielan The whole term has a second power. – Selena Dec 27 '18 at 09:26
  • @ParamanandSingh The whole term has a second power – Selena Dec 27 '18 at 09:26
  • $\sqrt{\pi}\sum_{a,b=1}^{100}ab\frac{\Gamma\left(\frac{a+b+1}{2}\right)}{\Gamma\left(\frac{a+b+2}{2}\right)}$ is approximately $5.68\cdot 10^6$. Reasonably the exercise is intended as an application of the central limit theorem: the integrand function, without the square, is well-approximated by $5050 \exp\left[-\frac{67}{2}\left(x-\pi/2\right)^2\right]$, so the value of the integral is close to $5050^2\sqrt{\frac{\pi}{67}}$. – Jack D'Aurizio Dec 27 '18 at 09:54
  • @Selena Yes, I noticed the whole term has a second power, which is why I wrote about the part inside of the brackets. Also, this is why I just said it might help as I was not sure myself and, as I stated at the end, it's likely not the best suggestion. – John Omielan Dec 27 '18 at 13:38

2 Answers2

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HINT: note that

$$\sum_{k=1}^m k y^k=y\sum_{k=1}^mky^{k-1}=y\sum_{k=1}^m\frac{d}{dy} y^k=y\frac{d}{dy}\sum_{k=1}^m y^k$$

Alternatively you can try to use the following formula to expand the square of the sum

$$\left(\sum_{k=1}^m k y^k\right)^2=\sum_{k=2}^{2m}c_k y^k$$

where

$$\begin{align}c_k:=&\sum_{j=1}^k j(k-j)=k\sum_{j=1}^k j-\sum_{j=1}^k j^2\\&=\frac12k^2(k+1)-\frac13(k+1)k(k-1)-\frac12 k(k+1)\\ &=(k+1)k\left(\frac{k}2-\frac{k-1}3-\frac12\right)\\ &=\frac16(k+1)k(k-1)\end{align}$$

However the value of the integral using this method would be a sum of $2m$ terms, that you would want to write in some compact form.

Masacroso
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  • Please explain more. Thanks. – Selena Dec 27 '18 at 09:20
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    @Selena using the first approach, using the formula for geometric sums, you get $$\frac{d}{dy}\sum_{k=1}^{100} y^k=\frac{d}{dy}\left(\frac{y^{101}-1}{y-1}-1\right)\=\frac{101 y^{100}(y-1)-y^{101}+1}{(y-1)^2}=\frac{100 y^{101}-101y^{100}+1}{(y-1)^2}$$ Hence $$\int_0^{\pi/2}\left(\sum_{k=1}^{100} k\sin^k x\right)^2, dx=\int_0^{\pi/2}\sin^2 x\frac{(100(\sin x)^{101}-101(\sin x)^{100}+1)^2}{(\sin x-1)^4}, dx$$ However the last integral seems not so easy to solve. Try the second approach with this – Masacroso Dec 27 '18 at 14:00
  • @Masacroso do you think that the form of $\int_{0}^{\frac{\pi}{2}}sin^zxdx$(as mentioned in the link) reduces to the same form as what I got below? If not, can you point out the problem? – Manraj Dec 27 '18 at 18:07
  • @Masacroso it seems I made an error here-"if n is odd, $I_n=\frac{(n-1)\pi}{2n}$ and $I_n=\frac{3(n-1)\pi}{8n}$" – Manraj Dec 27 '18 at 18:18
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$$S_n=\int_{0}^{\pi}[a_1sinx+a_2sin^2x+...+a_{n}sin^{200}x]dx$$ for some coefficients $a_1,a_2,...,a_n$.

Now let's look at the coefficient of $sin^{100}x$, this is all the sine terms multiplied such that the sum of the product is 100, eg. $1\times99,2\times98,...,99\times1$. Now you can observe that the coefficient must be of the form $\sum_{i=1}^{99}i(100-i)=\sum_{i=1}^{100}i(100-i)$(since the i=100 term is just zero). Similarly, for the kth power, you can write

$$a_k=\sum_{i=1}^{k}i(k-i)=\sum_{i=1}^{k}(ik-i^2)=k\times\frac{k(k+1)}{2}-\frac{k(k+1)(2k+1)}{6}=\frac{k(k+1)(3k-2k-1)}{6}=\frac{k(k^2-1)}{6}$$

So that $$S_n=\int_{0}^{\pi}[\sum_{n=1}^{200}\frac{n(n^2-1)}{6}sin^{n}x]dx$$

Now, let us evaluate$\int_{0}^{\pi}sin^{n}xdx$

$I_n=\int_{0}^{\pi}sin^{n}xdx=[-cosxsin^{n-1}x]_{0}^{\pi}+\int_{0}^{\pi}(n-1)sin^{n-2}xcosx\times cosxdx=\int_{0}^{\pi}(n-1)sin^{n-2}x(1-sin^2x)dx=(n-1)\int_{0}^{\pi}sin^{n-2}x-(n-1)\int_{0}^{\pi}sin^nxdx$

$$\Rightarrow I_n=(n-1)I_{n-2}-(n-1)I_n\Rightarrow I_n=\frac{(n-1)}{n}I_{n-2}$$ Also, you can check that $I_1=\frac{\pi}{2}(\int_{0}^{\pi}sin^2xdx=\int_{0}^{\pi}\frac{1-cos2x}{2}dx)$ and $I_2=\frac{3\pi}{8}(\int_{0}^{\pi}sin^4xdx=\int_{0}^{\pi}\frac{(1-cos2x)^2}{4}dx=\int_{0}^{\pi}\frac{(1+cos^22x+2cos2x)}{4}dx=\int_{0}^{\pi}\frac{(1+\frac{1+cos4x}{2}+cos2x)}{4}dx)$(I did this mainly using the result $sin^2x=\frac{1-cos2x}{2}\ and\ cos^2x=\frac{1+cos2x}{2}$ and the fact that $\int_{0}^{\pi}cos2nxdx=0=\int_{0}^{\pi}sin2nxdx,n\in\mathbb{N}$). So, if n is odd, $I_n=\frac{(n-1)\pi}{2n}$ and $I_n=\frac{3(n-1)\pi}{8n}$ if its even.

Now, you can use this in the original sequence to find the result which would give- $S_n=\sum_{n=1}^{100}\frac{2n(4n^2-1)\times 3(n-1)\pi}{6\times 8}+\sum_{n=1}^{100}\frac{(2n-1)(4n^2+1-2n-1)\times (n-1)\pi}{6\times 2}=\sum_{n=1}^{100}\frac{2n(2n-1)(n-1)\pi (3(2n+1)+8n-4)}{6\times 8}=\sum_{n=1}^{100}\frac{n(n-1)(2n-1)(14n-1)\pi}{24}$

Manraj
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