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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 :

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

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.

Note by Ronak Agarwal
1 year, 3 months ago

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Nice. And I really appreciate your patience while writing $$LATEX$$ · 1 year, 3 months ago

Thanks, how's the proof. · 1 year, 3 months ago

It's really nice . And since I am talking to jee topper and I am aspirant for JEE 2016 , I would like to ask the same question, which books do you referred to in Inorganic , Is it all about just remembering things up or analysing the concept of reaction mechanism ? · 1 year, 3 months ago

Can you explain how you got to that step after you wrote 'using this we have' in the first lemma? · 1 year, 2 months ago

Did you mean $$\Gamma(1)=1$$?

Also, your prove only shows that the Euler reflection formula is valid for $$x=\frac{1}{n}$$, where n is an integer more than 0.

Great prove though. · 1 year, 3 months ago

I know, I am trying to generalize this. · 1 year, 3 months ago

Isn't there a proof for lemma 1 which doesn't use induction? · 1 year, 3 months ago

Note that in Lemma 1 after multiplying both sides by $$\displaystyle \prod _{ i=1 }^{ n }{ (x-{ a }_{ i }) }$$ and subtracting $$1$$ from both sides, the polynomial generated on R.H.S. is of $$n-1$$ degrees, but has $$n$$ roots. Thus it is an identity. See Lagrange's Interpolation Polynomial. · 1 year, 3 months ago

Thanks! · 1 year, 3 months ago

It is obvious if you use covering lemma · 1 year, 2 months ago

You can try that for yourself. · 1 year, 3 months ago