# Thread: Today's calculation of integral #9

1. ## Today's calculation of integral #9

Find $\int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$

2. ## Ok

Originally Posted by Krizalid
Find $\int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$
You can rewrite it as $\int_0^1{\sum_{n=0}^{\infty}x^{n+2}}\cdot\bigg(-\sum_{n=0}^{\infty}\frac{x^{n}}{n}\bigg)^{-2}$...which then is equal to $\int_0^1{\sum_{n=0}^{\infty}x^{n+2}\cdot{2\sum_{n= 0}^{\infty}\sum_{n=0}^{\infty}\frac{n^2}{x^{2n}}}}$...and then $2\int_0^1{\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\s um_{n=0}^{\infty}\frac{x^{2-n}}{n}}$...so wed get $2\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\sum_{n=0}^ {\infty}\frac{x^{3-n}}{n(3-n)}$ evaluated from 0 to 1...I know I know I did it completely wrong...but will someone point out where? I just did this for novelty

3. Originally Posted by Mathstud28
I just did this for novelty
Then just don't do it if you don't know how to tackle what it comes.

It does exist a quick solution.

4. I did not solve it yet. Here is the first step I was thinking. Let $t = -\ln (1-x)$.
This gives us, $\int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt$.

5. Hello,

Let $u=\ln(1-x)$

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$du=-\frac{dx}{1-x}$

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$e^u=1-x$

$x=1-e^u$

$x^2=(1-e^u)^2$

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If x=0 then $u=0$

If x=1 then $u=-\infty$

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Inverting the extremities of the integral with the - sign coming from du

$\int_0^1\frac{x^2}{(1-x)\ln^2(1-x)} dx = \int_{-\infty}^0 \frac{(1-e^u)^2}{u^2} du$

Where did i go wrong ? :'(

6. Originally Posted by Moo
Where did i go wrong ? :'(
You did not solve it, look at my post above. This only changes the integral into a (probably) simpler one. Now you need to work with this new integral.

7. Actually, i've been trying but gave up an hour ago... The first parts of integration by parts were divergent oO

And ok, when i posted i saw the first version of your integral and got anxious because it was not the same...

8. ## So is it this?

Originally Posted by ThePerfectHacker
I did not solve it yet. Here is the first step I was thinking. Let $t = -\ln (1-x)$.
This gives us, $\int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt$.
is that then equal to $\int{\bigg(-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n-1}}{n!}+\sum_{n=0}^{\infty}\frac{(-2)^{n}x^{n-2}}{n!}+1\bigg)dx}$ $=-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot{n!}}+\sum_{n=0}^{\infty}\frac{ (-2)^{n}x^{n-1}}{(n-1)n!}+x$ evaluated from 0 to ∞?

9. Yop,

Where did $\frac{1}{t^2}$ go ? (it's it, instead of +1)

But your method doesn't simplify the problem... or i've missed a point ^^

10. ## oops

Originally Posted by Moo
Yop,

Where did $\frac{1}{t^2}$ go ? (it's it, instead of +1)

But your method doesn't simplify the problem... or i've missed a point ^^
what I said except change x to $\frac{-1}{x}$

11. Originally Posted by ThePerfectHacker
I did not solve it yet. Here is the first step I was thinking. Let $t = -\ln (1-x)$.
This is a good start.

Now find the other one.

12. Hint: construct a double integral. (Following on post #4.)

13. What we need is $\frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0.$ From there you may observe that the conclusion follows.

14. Originally Posted by Krizalid
What we need is $\frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0.$ From there you may observe that the conclusion follows.
Thank you for that. I had a similar idea: $\frac1{t^2} = \int_0^{\infty} e^{-t^2x} dx$ but that gave a nasty integral.

15. In general, $\frac{(n-1)!}{t^n} = \int_0^{\infty} x^{n-1} e^{-tx} dx$ for $t>0$ and $n\geq 1$. This seems to be a useful fact to know in double integration.

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