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If we have $f(w)= w^{-\frac{\alpha}{\beta}}\displaystyle\int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z $, what are the rules used to form $f(w)=\displaystyle\int_0^1 \frac{u^{\frac{\alpha}{\beta}-1}}{1-w u} \mathrm{d} u$?

Thanks a lot.

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3  
Let $z=wu$, $\mathrm dz=w\mathrm du$... –  J. M. Dec 6 '11 at 2:43
    
@J.M.: how about the integral? –  DRN Dec 6 '11 at 3:00
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The function inside the integral changes from $\displaystyle\frac{z^{\frac{\alpha}{\beta}-1}}{1-z}$ to $\displaystyle\frac{w^{\frac{\alpha}{\beta}-1}u^{\frac{\alpha}{\beta}-1}}{1-wu}$‌​. Putting all these together will do the job. –  Paul Dec 6 '11 at 3:08
    
@Paul: how about $\displaystyle\int_0^w$ changes to $\displaystyle\int_0^1 $ ? –  DRN Dec 6 '11 at 5:53
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Norlyda, I suggest you read about compositions, chain rule (derivative of a composed function) and the substitution theorem (for integrals, which is connected to the chain rule). –  AD. Dec 6 '11 at 6:56

1 Answer 1

up vote 2 down vote accepted

For fixed $w$, we consider $w^{-\frac{\alpha}{\beta}}\displaystyle\int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z$, which you called $f(w)$.

Let $z=wu$. Hence, if $z=w$, then $u=1$; if $z=0$, then if $z=0$, then $u=1$. Also, $\mathrm{d} z=w\,\mathrm{d}u$. On the other hand, $$\frac{z^{\frac{\alpha}{\beta}-1}}{1-z}=\frac{w^{\frac{\alpha}{\beta}-1}u^{\frac{\alpha}{\beta}-1}}{1-wu}.$$ Now putting all these calculations together, we obtain $$w^{-\frac{\alpha}{\beta}}\displaystyle\int_0^w \frac{z^{\frac{\alpha}{\beta}-1}}{1-z} \mathrm{d} z=w^{-\frac{\alpha}{\beta}}\int_0^1\frac{w^{\frac{\alpha}{\beta}-1}u^{\frac{\alpha}{\beta}-1}}{1-wu}\cdot w\,\mathrm{d}u=\int_0^1\frac{u^{\frac{\alpha}{\beta}-1}}{1-wu} \mathrm{d}u,$$ as required.

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thank you! i got it.. –  DRN Dec 6 '11 at 13:58
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Good! You may want to accept the answer if it helps you. –  Paul Dec 6 '11 at 14:21

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