need to solve integration

**CaptainBlack** · October 6th 2011, 02:49 AM

Originally Posted by Mhmh96

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Where does this "need" come from? Is that the original question? If not please post the real/original question.

CB

**Amer** · October 6th 2011, 03:18 AM

$t=\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2...}}}$

$t=\sqrt[x]{2t}\Rightarrow t^x = 2t \Rightarrow t = 2^{\frac{1}{1-x}}$

$2^{\frac{1}{1-x}} = e^{ln(2^{\frac{1}{1-x}})}= e^{\frac{1}{1-x}ln(2)}$

**Mhmh96** · October 6th 2011, 03:57 AM

The answer that gave to me here is not relevent to my problem

I need answer please !

**Mhmh96** · October 6th 2011, 03:58 AM

Thank you Amer i did not see your answer ^

**Prove It** · October 6th 2011, 06:23 AM

Amer's advice in post 4 gives you $\displaystyle \sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}} = e^{\frac{\ln{2}}{1 - x}}$ , so

$\displaystyle \begin{align*} \int{\frac{\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}}}{(1 - x)^2}\,dx} &= \int{\frac{e^{\frac{\ln{2}}{1 - x}}}{(1 - x)^2}\,dx} \\ &= \frac{1}{\ln{2}}\int{e^{\frac{\ln{2}}{1 - x}}\left[\frac{\ln{2}}{(1 - x)^2}\right]\,dx} \end{align*}$

Which now can be solved using the substitution $\displaystyle u = \frac{\ln{2}}{1 - x}$ .

In future I suggest you read and try to understand and use the help that you are given instead of telling these people that their advice is "not relevant" and demanding an answer.

**chisigma** · October 6th 2011, 06:45 AM

Originally Posted by Mhmh96

The 'core' of the question is the computation of...

$f(x)= \lim_{n \rightarrow \infty} a_{n}$ (1)

... where $a_{n}$ satisfies the difference equation...

$a_{n+1}= (2\ a_{n})^{\frac{1}{x}}$ (2)

The limit (1), if a finite limit exists, is the solution of the equation...

$\Delta_{n}= a_{n+1}-a_{n}= f(a_{n})=0$ (3)

... where...

$f(\xi)= (2\ \xi)^{\frac{1}{x}}-\xi$ (4)

With a little of patience You find that the solution of (3) is...

$\xi= 2^{- \frac{1}{1-x}}$ (5)

... so that is...

$\int \frac{2^{- \frac{1}{1-x}}}{(1-x)^{2}}\ dx = \frac{1}{\ln 2} 2^{-\frac{1}{1-x}} + c$ (6)

Kind regards

$\chi$ $\sigma$

**Amer** · October 6th 2011, 09:59 AM

Originally Posted by Amer

$t=\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2...}}}$

$t=\sqrt[x]{2t}\Rightarrow t^x = 2t \Rightarrow t = 2^{\frac{1}{1-x}}$

$2^{\frac{1}{1-x}} = e^{ln(2^{\frac{1}{1-x}})}= e^{\frac{1}{1-x}ln(2)}$

i made a mistake there is a minus sign $2^{\frac{-1}{1-x}} = e^{ln(2^{\frac{-1}{1-x}})}= e^{\frac{-1}{1-x}ln(2)}$

**Ackbeet** · October 6th 2011, 04:28 PM

Originally Posted by Mhmh96

The answer that gave to me here is not relevent to my problem

I need answer please !

Originally Posted by Prove It

Amer's advice in post 4 gives you $\displaystyle \sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}} = e^{\frac{\ln{2}}{1 - x}}$ , so

$\displaystyle \begin{align*} \int{\frac{\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}}}{(1 - x)^2}\,dx} &= \int{\frac{e^{\frac{\ln{2}}{1 - x}}}{(1 - x)^2}\,dx} \\ &= \frac{1}{\ln{2}}\int{e^{\frac{\ln{2}}{1 - x}}\left[\frac{\ln{2}}{(1 - x)^2}\right]\,dx} \end{align*}$

Which now can be solved using the substitution $\displaystyle u = \frac{\ln{2}}{1 - x}$ .

In future I suggest you read and try to understand and use the help that you are given instead of telling these people that their advice is "not relevant" and demanding an answer.

The OP'er was likely referring to some deleted posts of chisigma's.

**chisigma** · October 6th 2011, 10:28 PM

Originally Posted by Ackbeet

The OP'er was likely referring to some deleted posts of chisigma's.

I confirm that!... my first approach was fully wrong and I apologize for that!

...

Kind regards

$\chi$ $\sigma$

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