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Here is my attempt:

$$\begin{split}\int_{0}^{1}\frac{x^{a-1}}{x+1}\ dx&=\sum_{k=0}^{\infty }(-1)^k\int_{0}^{1}\ x ^{a+k-1}dx\\ &=\sum_{k=0}^{\infty }\frac{(-1)^k}{a+k}\\ &=\sum_{k=0}^{\infty }\left [ \frac{1}{2k+a} -\frac{1}{2k+1+a}\right ]\\ &=\frac{1}{2}\sum_{k=1}^{\infty }\left [ \frac{1}{k+\frac{a-2}{2}}-\frac{1}{k+\frac{a-1}{2}} \right ]\\ &=\frac{1}{2}\left [ \Psi \left (1+\frac{a-1}{2} \right ) -\Psi \left (1+\frac{a-2}{2} \right )\right ]\\ &=\frac{1}{2}\left [ \Psi \left (\frac{a+1}{2} \right )-\Psi \left (\frac{a}{2} \right ) \right ] \end{split}$$

Can we compute the integral above in different methods?

All approaches are highly appreciated.

Simply Beautiful Art
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Ahmed Hejazi
  • 559

2 Answers2

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Note: I was trying to do something else but it didn't work out like I thought it would, but i thought I would ost it anyways.

My answer isn't too useful but it is interesting. My answer uses a little differentiation under the integral sign. My answer is a complicated way of doing it your way... But it was fun!

If anything is unclear in my answer, please ask and Me or someone else will answer.

$ \begin{array}{l} \int\limits ^{1}_{0}\frac{x^{a-1}}{x+1} dx=\int\limits ^{1}_{0}\frac{x^{a} x^{-1}}{x+1} dx=f( a) \Longrightarrow \\ f'( a) =\int\limits ^{1}_{0}\frac{ln( x) x^{a} x^{-1}}{x+1} dx,\ x=e^{-t} ,\ dx=-e^{-t} dt,\ t( 0) =\infty ,\ t( 1) =0\Longrightarrow \\ \int\limits ^{1}_{0}\frac{ln( x) x^{a} x^{-1}}{x+1} dx=-\int\limits ^{\infty }_{0}\frac{te^{-ta}}{1-\left( -e^{-t}\right)} dt=-\int\limits ^{\infty }_{0} te^{-ta}\sum ^{\infty }_{n=1}\left[\left( -e^{-t}\right)^{n-1}\right] dt=-\sum ^{\infty }_{n=1}\left[( -1)^{n-1}\int\limits ^{\infty }_{0} te^{-ta} e^{t( 1-n)} dt\right]\\ =\sum ^{\infty }_{n=1}\left[( -1)^{n}\int\limits ^{\infty }_{0} te^{-t( n-1+a)} dt\right] ,\ n\in \mathbb{N} ,\ \Re ( a) >0\Longrightarrow n-1+a\ is positive\\ \therefore \lim _{t\rightarrow \infty }\left( e^{-t( n-1+a)}\right) =0\Longrightarrow \\ \\ \sum ^{\infty }_{n=1}\left[( -1)^{n}\int\limits ^{\infty }_{0} te^{-t( n-1+a)} dt\right] =\sum ^{\infty }_{n=1}\left[( -1)^{n}\left(\lim\limits ^{ \begin{array}{l} t=\infty \\ \end{array}}_{t=0}\left(\frac{te^{-t( n-1+a)}}{n-1+a}\right) -\frac{1}{1-n-a}\int\limits ^{\infty }_{0} e^{-t( n-1+a)} dt\right)\right]\\ =\sum ^{\infty }_{n=1}\left[( -1)^{n}\left(\frac{1}{( 1-n-a)^{2}}\right)\right]\\ \\ f'( a) =\sum ^{\infty }_{n=1}\left[\frac{( -1)^{n}}{( 1-n-a)^{2}}\right]\\ \therefore f( a) =\sum ^{\infty }_{n=1}\left[( -1)^{n}\int \frac{1}{( 1-n-a)^{2}} da\right]\\ \\ \int \frac{1}{( 1-n-a)^{2}} da,\ t=1-n-a,\ a=1-n-t\Longrightarrow da=-1dt\Longrightarrow \\ \int \frac{1}{( 1-n-a)^{2}} da=-\int t^{-2} da=\frac{1}{t} =\frac{1}{( 1-n-a)} +C\\ \therefore f( a) =\sum ^{\infty }_{n=1}\left[\frac{( -1)^{n}}{( 1-n-a)}\right] =\sum ^{\infty }_{n=0}\left[\frac{( -1)^{n-1}}{-n-t}\right] =\sum ^{\infty }_{n=0}\left[\frac{( -1)^{n}}{n+t}\right] =\Phi ( -1,\ 1,\ a) \Longrightarrow \\ \\ \int\limits ^{1}_{0}\frac{x^{a-1}}{x+1} dx=\Phi ( -1,\ 1,\ a)\\ \\ in\ case\ you\ don't\ know:\\ \\ \Phi ( z,\ s,\ \alpha ) =\sum ^{\infty }_{n=0}\left[\frac{z^{n}}{( n+\alpha )^{s}}\right]\\ \\ \frac{a}{1-r} =\sum ^{\infty }_{n=1} ar^{n-1} ,\ | r| < 1,\ \end{array}$

Highvoltagemath
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A substitution of $u=x+1$ makes the process much easier using the binomial expansion on the top.

Mohammad Riazi-Kermani
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