# Thread: need to solve integration

2. ## Re: need to solve integration

Originally Posted by Mhmh96
Where does this "need" come from? Is that the original question? If not please post the real/original question.

CB

3. ## Re: need to solve integration

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

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

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

4. ## Re: need to solve integration

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

6. ## Re: need to solve integration

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.

7. ## Re: need to solve integration

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

$\chi$ $\sigma$

8. ## Re: need to solve integration

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

9. ## Re: need to solve integration

Originally Posted by Mhmh96
The answer that gave to me here is not relevent to my problem

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.

10. ## Re: need to solve integration

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

$\chi$ $\sigma$