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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 :

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.

Note by Ronak Agarwal
9 months, 3 weeks ago

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Nice. And I really appreciate your patience while writing \(LATEX\) Pranjal Prashant · 9 months, 3 weeks ago

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@Pranjal Prashant Thanks, how's the proof. Ronak Agarwal · 9 months, 3 weeks ago

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@Ronak Agarwal 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 ? Pranjal Prashant · 9 months, 3 weeks ago

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Can you explain how you got to that step after you wrote 'using this we have' in the first lemma? Donkey Kong · 8 months, 2 weeks ago

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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. Julian Poon · 9 months, 3 weeks ago

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@Julian Poon I know, I am trying to generalize this. Ronak Agarwal · 9 months, 3 weeks ago

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Isn't there a proof for lemma 1 which doesn't use induction? Aditya Kumar · 9 months, 3 weeks ago

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@Aditya Kumar 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. Ishan Singh · 9 months, 3 weeks ago

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@Ishan Singh Thanks! Aditya Kumar · 9 months, 2 weeks ago

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@Aditya Kumar It is obvious if you use covering lemma Praneeth Kacham · 9 months, 1 week ago

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@Aditya Kumar You can try that for yourself. Ronak Agarwal · 9 months, 3 weeks ago

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Great! Can you put that inside here? Pi Han Goh · 9 months, 3 weeks ago

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