Proof of euler reflection formula

Here's another proof of Euler reflection formula :

$Lemma1\quad :$

$P(n) : \displaystyle \prod _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } }$ for all $n$ belonging to $N$

Proof :

I will prove it by induction :

It can be easily verified that $P(1)$ is true :

Assume $P(n)$ is true :

than $\displaystyle \prod _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } }$

Multiplying both sides with $\dfrac{1}{(x-{a}_{n+1})}$ we have :

$\displaystyle \prod _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } (\dfrac { 1 }{ (x-{ a }_{ n+1 }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } ) } }$

It is easy to check that :

$\dfrac { 1 }{ (x-{ a }_{ i }) } \dfrac { 1 }{ (x-{ a }_{ n+1 }) } =\dfrac { 1 }{ ({ a }_{ i }-{ a }_{ n+1 }) } \dfrac { 1 }{ (x-{ a }_{ i }) } +\dfrac { 1 }{ ({ a }_{ n+1 }-{ a }_{ i }) } \dfrac { 1 }{ (x-{ a }_{ n+1 }) }$

Using this we have :

$\displaystyle \prod _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } +\dfrac { 1 }{ (x-{ a }_{ n+1 }) } \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ { (a }_{ n+1 }-{ a }_{ i }) } } \prod _{ j=1, j \neq i }^{ n }{ \dfrac { 1 }{ { a }_{ i }-{ a }_{ j } } }$

Now using P(n) we can say that :

$\displaystyle \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ { (a }_{ n+1 }-{ a }_{ i }) } } \prod _{ j=1,j\neq i}^{ n }{ \dfrac { 1 }{ { a }_{ i }-{ a }_{ j } } } =\prod _{ i=1 }^{ n }{ \dfrac { 1 }{ ({ a }_{ n+1 }-{ a }_{ i }) } } =\prod _{ i=1,i\neq n+1\quad }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ n+1 }-{ a }_{ i }) } }$

Using this in our upper result we have :

$\displaystyle \prod _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } } =\sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } +\dfrac { 1 }{ (x-{ a }_{ n+1 }) } \prod _{ j=1,j\neq n+1 }^{ n+1 }{ \dfrac { 1 }{ { a }_{ n+1 }-{ a }_{ j } } } =\sum _{ i=1 }^{ n+1 }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n+1 }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } }$

Hence $P(n)$ is true implies $P(n+1)$ is true implies $P(n)$ is true for all $n$ belongs to $N$

$Lemma2 \quad :$

$\displaystyle \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } =\dfrac { 1 }{ P'({ a }_{ i }) }$, where $P(x)$ is a monic polynomial having roots {${a}_{k}$}, for $1 \leq k \leq n$

Proof : $\displaystyle P(x) = \prod _{j=1}^{ n }{ (x-{ a }_{ j }) }$

Dividing both sides by $(x-{a}_{i})$ and taking limit $x \rightarrow {a}_{i}$ we have :

$\displaystyle \lim _{ x\rightarrow { a }_{ i } }{ \dfrac { P(x) }{ x-{ a }_{ i } } } =\prod _{ j=1,j\neq i }^{ n }{ ({ a }_{ i }-{ a }_{ j }) }$

Applying L-Hopitals rule we have our result :

Now we start with the integral :

$\displaystyle I(n) = \int _{ 0 }^{ \infty }{ \dfrac { dx }{ 1+{ x }^{ n } } }$

Let it's roots be {${a}_{k}$}, for $1 \leq k \leq n$

Then we have :

$\displaystyle I(n) = \int _{ 0 }^{ \infty }{ \prod _{ j=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ j }) } }dx }$

Applying lemma 1 we have :

$\displaystyle I(n) = \int _{ 0 }^{ \infty }{ \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \prod _{ j=1,j\neq i }^{ n }{ \dfrac { 1 }{ ({ a }_{ i }-{ a }_{ j }) } } } dx }$

Applying lemma 2 we have :

$\displaystyle I(n) = \int _{ 0 }^{ \infty }{ \sum _{ i=1 }^{ n }{ \dfrac { 1 }{ (x-{ a }_{ i }) } \dfrac { 1 }{ n{ a }_{ i }^{ n-1 } } } dx }$

Using the fact that ${a}_{i}^{n}=-1$ we have :

$\displaystyle -\dfrac { 1 }{ n } \int _{ 0 }^{ \infty }{ \sum _{ i=1 }^{ n }{ \dfrac { { a }_{ i } }{ (x-{ a }_{ i }) } } dx }$

Changing the order of integration and summation and integrating it we have :

$\displaystyle I(n) = \dfrac { 1 }{ n } \lim _{ p\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ { a }_{ i }ln(x-{ a }_{ i }) } { | }_{ p }^{ 0 } }$

$\displaystyle I(n) = \dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln(-{ a }_{ i }) } -\lim _{ p\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ ln(p-{ a }_{ i }) } } )$

Now $\displaystyle \lim _{ p\rightarrow \infty }{ \sum _{ i=1 }^{ n }{ { a }_{ i }ln(p-{ a }_{ i }) } } =\lim _{ p\rightarrow \infty }{ ln(p)\sum _{ i=1 }^{ n }{ { a }_{ i } } -\sum _{ i=1 }^{ n }{ { a }_{ i }ln(1-\dfrac { { a }_{ i } }{ p } ) } } =0$

Here we have used the fact that sum of roots is 0 and applied the limit.

Now the value of our integral becomes :

$\displaystyle I(n) = \dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln(-{ a }_{ i }) } )=\dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln({ a }_{ i }) } +ln(-1)\sum _{ i=1 }^{ n }{ { a }_{ i } } )=\dfrac { 1 }{ n } (\sum _{ i=1 }^{ n }{ { a }_{ i }ln({ a }_{ i }) } )$

Using the general form of root $\large {a}_{k} = { e }^{ \left({ \frac { (2k-1)\pi i }{ 2 } }\right) }$

Our integral becomes :

$\displaystyle I(n) = \dfrac { 1 }{ n } \left(\sum _{ k=1 }^{ n }{ \frac { (2k-1)\pi i }{ n } { e }^{ \frac { (2k-1)\pi i }{ n } } } \right)= \large \dfrac { 2\pi i }{ { n }^{ 2 }{ e }^{ \frac { \pi i }{ n } } } \sum _{ k=1 }^{ n }{ k{ e }^{ \frac { 2\pi ik }{ n } } }$

Here I have again used the fact that sum of roots is 0.

Using the concept of Arithmetic-Geometric series it can be easily shown that :

$\displaystyle \sum _{ k=1 }^{ n }{ k{ x }^{ k } } =\dfrac { x({ x }^{ n }-1) }{ { (1-x) }^{ 2 } } -\dfrac { n{ x }^{ n+1 } }{ 1-x }$

Using this we have :

$I(n) = \large \dfrac { 2\pi i }{ n{ e }^{ \frac { \pi i }{ n } } } \left(\dfrac { { e }^{ \frac { 2\pi i }{ n } } }{ { e }^{ \frac { 2\pi i }{ n } }-1 } \right)$

Simplifying it we get :

$\displaystyle I(n) = \dfrac { \pi }{ n } \csc { \dfrac { \pi }{ n } }$

Again back to our integral :

$\displaystyle I(n) = \int _{ 0 }^{ \infty }{ \dfrac { dx }{ 1+{ x }^{ n } } }$

Use the substitution $y = \dfrac { dx }{ 1+{ x }^{ n } }$, we have :

$\displaystyle I(n) = \dfrac { 1 }{ n } \int _{ 0 }^{ 1 }{ { y }^{ \frac { -1 }{ n } }{ (1-y) }^{ \frac { 1 }{ n } -1 }dy }$

By defintion of beta function this evaluates to :

$\displaystyle I(n) = \dfrac { 1 }{ n } \dfrac { \Gamma \left(\dfrac { 1 }{ n } \right)\Gamma \left(1-\dfrac { 1 }{ n } \right) }{ \Gamma (1) }$

Comparing our two results we have :

$\dfrac { 1 }{ n } \Gamma \left(\dfrac { 1 }{ n } \right)\Gamma \left(1-\dfrac { 1 }{ n } \right)=\dfrac { \pi }{ n } \csc { \dfrac { \pi }{ n } }$

I have used $\Gamma(1)=1$

Finally we have :

$\Gamma \left(\dfrac { 1 }{ n } \right)\Gamma \left(1-\dfrac { 1 }{ n } \right)=\pi \csc { \dfrac { \pi }{ n } }$

Take $\dfrac{1}{n} = x$ to get :

$\Gamma (x)\Gamma (1-x)=\pi \csc { \pi x }$

$\LARGE HENCE \quad PROVED$

Sorry for the fact that the proof looks too messy, and that I have jumped many steps. It's largely based on manipulations on complex numbers.