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From a journal, they proved this equality:

$$ \frac{z}{\alpha -1}\left(\int_0^1 \frac{t^{\frac{1}{\alpha}}}{1-tz} dt -\alpha \int_0^1 \frac{v}{1-vz} dv\right) = \int_0^1 t^{\frac{1}{\alpha}} \left(\int_0^1 \frac{vz}{1-tvz}dv\right)dt $$

I already tried various methods to prove it but I can't get the same equality. Do you have any idea? I really need to refer this journal to assist my research. Thank you.

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Your previous questions have kept referring to some journal article, but it seems you haven't mentioned which one... –  J. M. Dec 12 '11 at 3:39
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Just say the title of the article, and in which journal it was published. It is common here to give citations, you know... –  J. M. Dec 12 '11 at 4:04
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You say it's proved in the journal, but can't prove it? Doesn't the proof in the journal work? –  Bruno Joyal Dec 12 '11 at 4:10
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Because my supervisor bought that journal, so I thought that it is not available to the public... Care to explain? –  Did Dec 12 '11 at 9:30

2 Answers 2

The purported identity is clearly seen wrong. Consider $\alpha = \frac{1}{2}$ and $z = \frac{1}{2}$. Then: $$ I_1 = \int_0^1 \frac{t^{1/\alpha}}{1- t z} \mathrm{d} t = \int_0^1 \frac{t^2}{1-t/2} \mathrm{d} t = \int_0^1 \left( \frac{8}{2 - t} - 4 - 2 t \right) \mathrm{d} t = 8 \cdot \log 2 - 5 $$ $$ I_2 = \int_0^1 \frac{v}{1- v z} \mathrm{d} v = \int_0^1 \frac{v}{1-v/2} \mathrm{d} v = \int_0^1 \left( \frac{4}{2-v} -2 \right) \mathrm{d} v = 4 \cdot \log 2 - 2 $$ $$ \begin{eqnarray} I_3 &=& \int_0^1 \int_0^1 t^{1/\alpha} \left( \frac{v z}{1- t v z} \mathrm{d} v \right) \mathrm{d} t = \int_0^1 \int_0^1 t^2 \left( \frac{v z}{1- t v z} \mathrm{d} v \right) \mathrm{d} t = \int_0^1 \int_0^1 \left( -t + \frac{t}{1-t v/2} \right) \mathrm{d} v \, \mathrm{d} t \\ &=& \int_0^1 \left( -t - 2 \cdot \log(1-t/2) \right) \mathrm{d} t = \frac{3}{2} - 2 \cdot \log 2 \end{eqnarray} $$ But $$ \frac{z}{\alpha -1} \left( I_1 - \alpha I_2 \right) = \frac{1}{2} I_2 - I_1 = 4 - 6 \cdot \log 2 \not= I_3 = \frac{3}{2} - 2 \cdot \log 2 $$

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@sarah: could you give me some idea? I also found that someone ask similar question in this link –  DRN Dec 12 '11 at 14:38

If $|z|\lt1$, expanding both sides as power series in $z$ yields that, for every $n\geqslant1$, the coefficients of $z^n$ in the LHS and in the RHS are $$ \frac1{\alpha -1}\left(\int_0^1 t^{1/\alpha+n-1} dt -\alpha \int_0^1 v^n dv\right)\quad\text{and}\quad\int_0^1 t^{1/\alpha} \left(\int_0^1 t^{n-1}v^ndv\right)dt. $$ These are $$ \frac1{\alpha -1}\left(\frac1{1/\alpha+n} -\alpha\frac1{n+1}\right)\quad\text{and}\quad\frac1{n+1}\int_0^1 t^{1/\alpha+n-1}dt, $$ that is, $$ \frac{-\alpha n}{(n+1)(1+\alpha n)}\quad\text{and}\quad\frac{\alpha n}{(n+1)(1+\alpha n)}. $$ Hence the LHS is the RHS times $(-1)$.

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