Writing recurrence relations for $I_n=\int{\frac{1}{\sin^n{x}}}dx$ in sigma notation

Question

I have calculated the reduction formula for: $$I_n=\int{\frac{1}{\sin^n{x}}}dx$$ as: $$I_n=\frac{-\csc^{n-2}({x})\cot({x})+(n-2)I_{n-2}}{n-1}$$.

How do I evaluate $I_{2m}$ and $I_{2m+1}$? The text says that the answer must be given in summation notation.

Answer 1

(Hint) As suggested in a comment, start by writing out one or more specific examples:

$$I_4=\frac{-\csc^2(x)\cot(x)+2I_2}{3}\\ =\frac{-\csc^2(x)\cot(x)}3-\frac{2\cot(x)+0}3$$

$$I_3=\frac{-\csc(x)\cot(x)+I_1}2\\ =\frac{-\csc(x)\cot(x)}2+\frac 12\int\frac{\text dx}{\sin x}$$

It should be clear from the above why the sums for $I_{2m},I_{2m+1}$ are separated. By the reduction being on a cycle of $2$ a non-factorial denominator should be expected, although perhaps the odd-index integral could make use of the factorial with a helper term.

Answer 2

Integrate $I_n=\int \frac1{\sin^n x}\ dx$ in summation forms as follows: \begin{align} I_{2m}=& -\int (1+\cot^2 x)^{m-1}d(\cot x)\>\>\>\>\>\>\> t=\cot x\\=& -\int (1+t^2)^{m-1}dt=-\sum_{k=0}^{m-1}\binom {m-1}k\int t^{2k}dt =-\sum_{k=0}^{m-1}\binom {m-1}k\frac{t^{2k+1}}{2k+1}\\ \\ I_{2m+1}=& -\int \frac{1}{(1-\cos^2 x)^{m+1}} d(\cos x)\>\>\>\>\>\>\> \cos x=\frac{1-t}{1+t}\\ =& -\frac1{2^{2m+1}}\int \frac{(1+t)^{2m}}{t^{m+1}}dt = -\frac1{2^{2m+1}}\sum_{k=0}^{2m}\binom {2m}k\int t^{k-m-1}dt\\ = & -\frac1{2^{2m+1}}\bigg\{\binom {2m}m\ln t+\sum_{k=0}^{m-1}\bigg[\binom {2m}k\frac{t^{k-m}}{k-m} + \binom {2m}{k+m+1}\frac{t^{k+1}}{k+1} \bigg]\bigg\}\\ \end{align}

Answer 3

Just in case you want the explicit representation for this integral, in a paper I'm writing covering Nonlocal curvature, this integral comes up a lot: $$\int_{\tfrac{\pi}{2}}^{x}\csc^n(x)dx=-\tfrac{1}{2}\Big(\tfrac{\cos x}{\left|\cos x\right|}\Big)\beta_{\cos^2(x)}\Big(\tfrac{1}{2},\tfrac{1-n}{2}\Big).$$ Provided that $n\in\mathbb{R},|n|\leq{}1,0\leq{}x\leq{}\pi$. The trick comes in once we realize that the indefinite integral can be expressed as $$\int_a^b\csc^n(x)dx=\int_{\pi/2}^b\csc^n(x)dx-\int_{\pi/2}^a\csc^n(x)dx.$$

[Proof] Set the previously defined integral to the variable J. Because $x\in[0,\pi]$, we have: $$J=\frac{\cos(x)}{|\cos(x)|}\int_{\frac{\pi}{2}}^{x}\frac{\sin(\theta)}{\left|\sin^{n+1}(\theta)\right|}\left(\frac{\cos(\theta)}{|\cos(\theta)|}\right) \ \mathrm{d\theta}.$$ Recall that the absolute value of a function is defined as the square root of that function squared. Applying this identity to our integral gives us: $$J=\frac{\cos(x)}{2|\cos(x)|}\int_{\frac{\pi}{2}}^{x}\frac{2\sin(\theta)\cos(\theta)}{\sqrt{\cos^2(\theta)\sin^{2n+2}(\theta)}} \ \mathrm{d\theta}.$$ Applying the substitution $u=\cos^2(x)$ yields the following result: $$J=-\frac{1}{2}\left ( \frac{\cos(x)}{\left|\cos(x)\right|} \right )\int_0^{\cos^2(x)}u^{\left(\frac{1}{2}\right)-1}(1-u)^{\left(\frac{1-n}{2}\right)-1}\ \mathrm{du}.$$ Recalling the definition of the incomplete beta function $\beta_z(a,b)$ $$\beta_z(a,b)=\int_0^{z}u^{a-1}(1-u)^{b-1}\ \mathrm{du}$$ leads one to conclude that $$J=-\frac{1}{2}\left(\frac{\cos(x)}{\left|\cos(x)\right|}\right)\beta_{\cos^2(x)}\left ( \frac{1}{2},\frac{1-n}{2}\right).$$ This may aid in finding the eventual recursion relation... Good luck!