On the Value Of an Integral

I would like to show you a proof for the integral, 01xb1(1x)cb1ln(x)dx\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx It is for the proving of my previous discussion,"On the Summation of Hypergeometric series". Observe, ddb(B(b,cb))=01xb1(1x)cb1(ln(x)ln(1x))dx\frac{d}{db}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}(\ln (x)-\ln (1-x))dx Then by splitting the integral we get, ddb(B(b,cb))=01xb1(1x)cb1ln(x)dx01xb1(1x)cb1ln(1x)dx\frac{d}{db}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx-\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (1-x)dx We have our integral but with an additional integral,for that lets use this integral.Since, B(b,cb)=01xb1(1x)cb1dx\Beta(b,c-b)=\int_0^1x^{b-1}(1-x)^{c-b-1}dx Then taking d.w.r.c,we get, ddc(B(b,cb))=01xb1(1x)cb1ln(1x)dx\frac{d}{dc}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (1-x)dx And hey! That's our required equation.So, ddb(B(b,cb))+ddc(B(b,cb))=01xb1(1x)cb1ln(x)dx\frac{d}{db}(\Beta(b,c-b))+\frac{d}{dc}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx So the only thing left is to find the value of the derivative,which is quite easy once you have the knowledge of beta function and the diagamma function, ddb(B(b,cb))=B(b,cb)(ψ(b)ψ(cb)),ddc(B(b,cb))=B(b,cb)(ψ(cb)ψ(c))\frac{d}{db}(\Beta(b,c-b))=\Beta(b,c-b)(\psi(b)-\psi(c-b)),\frac{d}{dc}(\Beta(b,c-b))=\Beta(b,c-b)(\psi(c-b)-\psi(c)) Then adding both derivative we have, 01xb1(1x)cb1ln(x)dx=B(b,cb)(ψ(b)ψ(c))\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx=\Beta(b,c-b)(\psi(b)-\psi(c)) And as our corollary ,(I love corollary), 01xb1(1x)bln(x)dx=πsin(πb)(ψ(b)+γ)\int_0^1x^{b-1}(1-x)^{-b}\ln (x)dx=\frac{\pi}{\sin(\pi b)}(\psi(b)+\gamma) Where γ\gamma is Euler constant about 0.577.

Note by Aruna Yumlembam
2 weeks, 6 days ago

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