Krdytkyu

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

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

$$S_n=int_{0}^{pi}[a_1sinx+a_2sin^2x+...+a_{n}sin^{200}x]dx$$ 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. $1times99,2times98,...,99times1$. 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

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

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$int_{0}^{pi}sin^{n}xdx$

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

$$Rightarrow I_n=(n-1)I_{n-2}-(n-1)I_nRightarrow I_n=frac{(n-1)}{n}I_{n-2}$$
Also, you can check that $I_1=frac{pi}{2}(int_{0}^{pi}sin^2xdx=int_{0}^{pi}frac{1-cos2x}{2}dx)$ and $I_2=frac{3pi}{8}(int_{0}^{pi}sin^4xdx=int_{0}^{pi}frac{(1-cos2x)^2}{4}dx=int_{0}^{pi}frac{(1+cos^22x+2cos2x)}{4}dx=int_{0}^{pi}frac{(1+frac{1+cos4x}{2}+cos2x)}{4}dx)$(I did this mainly using the result $sin^2x=frac{1-cos2x}{2} and cos^2x=frac{1+cos2x}{2}$ and the fact that $int_{0}^{pi}cos2nxdx=0=int_{0}^{pi}sin2nxdx,ninmathbb{N}$). 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-
$S_n=sum_{n=1}^{100}frac{2n(4n^2-1)times 3(n-1)pi}{6times 8}+sum_{n=1}^{100}frac{(2n-1)(4n^2+1-2n-1)times (n-1)pi}{6times 2}=sum_{n=1}^{100}frac{2n(2n-1)(n-1)pi (3(2n+1)+8n-4)}{6times 8}=sum_{n=1}^{100}frac{n(n-1)(2n-1)(14n-1)pi}{24}$

HINT: note that

$$sum_{k=1}^m k y^k=ysum_{k=1}^mky^{k-1}=ysum_{k=1}^mfrac{d}{dy} y^k=yfrac{d}{dy}sum_{k=1}^m y^k$$

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

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

where

$$begin{align}c_k:=&sum_{j=1}^k j(k-j)=ksum_{j=1}^k j-sum_{j=1}^k j^2\&=frac12k^2(k+1)-frac13(k+1)k(k-1)-frac12 k(k+1)\
&=(k+1)kleft(frac{k}2-frac{k-1}3-frac12right)\
&=frac16(k+1)k(k-1)end{align}$$

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.