## Monday 16 January 2012

### Generating functions part 2: by integration on log series

Last time we have talked about proving identities through differentiation on known identities, so this time we will deal with identities by integration, mainly on the log series. We know that calculus is the bridge between rational functions and logarithm series. We can transform series of rational expression into geometric sum, then by integration we can express logarithm series in terms of infinite sum of polynomials.

Example. show that $\sum^{n}_{i=1}\frac{1}{i(i+1)}=\frac{n}{n+1}$
Observe that $\int \frac{1}{x+x^2}dx=\ln x-\ln (x+1)+C$
(I don't expect readers to have the ability to compute this one because it involve a very tricky integral $\int \csc x dx$ which is pretty hard.)
Now $\sum [\ln x-\ln (x+1)+C_i]=C-\ln (x+1)$ where C is a constant.
Differentiate both sides ones, $\sum \frac{1}{x(x+1)}=C-\frac{1}{1+x}$, by checking the first term, C=1 so $\sum \frac{1}{x(x+1)}=\frac{n}{n+1}$, Q.E.D..

Example. Show that $\ln (1+x)=\sum \frac{-(-x)^n}{n}$ for $x\in [-1,1]$.

In traditional approach this proof involves limit of exponential series and tricky substitution but with this integration approach we can make it easy.

Since $\frac{1}{1-x}=\sum x^r$ for $x\in [-1,1]$, integration gives $\int \frac{1}{1\pm x}=\ln (1\pm x)+C$, we have $\ln (1-x)+C=\sum \frac{x^{r+1}}{r+1}$ for $x\in [-1,1]$. Now we solve the C and we get zero. Q.E.D..
Note that the original summation starts with the term 1, after integration it starts with the term x.

Before this one ends, we've got a technical analytical problem: how to determine C in the summation term after integration?
Let's say $\sum f(x)=g(x)$. After integration it's like $\sum(F(x)+C(x))=G(x)+C'(x)$. Since $\frac{d(G(x)+C'(x))}{dx}=g(x)$, C'(x) is a constant function. If LHS is a finite sum, i.e., it's in the form $\sum^{n}f(x)=g(n)$, by considering $F(n)=\sum^{n}[F(i)+C(i)]-\sum^{n-1}[F(i)-C(i)]$ we have $\Delta C(n)=0$, so C(n) is a zero function. The constant from integration comes from the first term. If it's infinite sum like the second example, we write $G(x)+C=\sum [F(x)+C'(x)]$, differentiate RHS once should give f(x), so C(x) is a constant function. However since it's an infinite sum. If it's non-zero LHS diverges which trivially a contradiction. Therefore the constant term in the summation is zero as well. How about the constant term on the LHS, must it be zero? It's left for readers to explore themselves.

Exercise:
1a) Show that $\frac{1}{2}\ln \frac{1+x}{1-x}=\sum^{\infty}_{i=1} \frac{(2i-1)^{2i-1}}{2i-1}$.
b) Hence show that $\frac{1}{2}\ln (1+\frac{1}{n})=\sum^{\infty}_{i=1}\frac{1}{(2i-1)(2n+1)^{2i-1}}$
c) By (a) show that the two infintie sums are equal: $\sum \frac{1}{2ix^{2i}},2\sum \frac{1}{(2i-1)(2x^2-1)^{2i-1}}$.
2) If $\phi \neq n\pi$, show that $\ln \csc \phi=\sum^{\infty}_{i=1}\frac{\cos ^{2i}\phi}{2i}$
3) Prove or disprove the existance of non-zero constant term derived from the integration of known identities.
4) In the second example show that C=0.