The Proof is Trivial!

...

$= \left[x\arcsin \left( \frac{x}{1-x} \right) \right]_0^{1/2} - \int_0^{1/2} \frac{x}{\sqrt{1-2x}}dx$ (*)

I left out the sub and rearrangement. I completely understand why LOTF does it now. Ask if you want.

I used IBP as well. Couple of questions/comments:

1) On the line I've added (*) to, haven't you missed out a factor of (1-x) on the denominator.
2) Whilst you might not want to write out the entire sub, just stating which substitution you've used (In this case $u^2=1-2x$ ) might be nice for readers.

(Also, I just realised - I forgot that the integral should have a factor of 2 added to it (From the initial uniform distribution of x) so the answer is actually supposed to be double what I stated. :O )

I'll fix that, It's because I made a mistake when I did it on the whiteboard and copied the wrong things down

Okay

I did wonder, because there are two halfs? I wasn't confident enough with my reasoning to say though.

It's because you have to take the shorter half, so it's uniformly distributed between 0 and 1/2. The probability density function of this is therefore 2, which gives an extra factor of 2 when doing the integral. The problem was, I moved it to the outside and then had all the integration to do, but forgot to move it back in by the time the integration was done.

Yeh, yeh that's what I meant. It was the initial " x<0.5 malarkey" I had to think about most. For some reason stats just doesn't click with me at all.

Problem 202: *
Just a bit of fun. Really not very difficult, but it looks nice
Find:
$\displaystyle\int \dfrac{1}{x\ln(x)\ln(\ln(x))} \ dx$

And hence evaluate:

$\displaystyle\int^{e^{e^e}}_{e^e} \dfrac{1}{x\ln(x)\ln(\ln(x))} \ dx$

Solution 202:

$Let y = lnx$

$dy = \dfrac{1}{x} \ dx$

$\Rightarrow \displaystyle\int \dfrac{1}{x\ln(x)\ln(\ln(x))} \ dx = \displaystyle\int \dfrac{1}{ylny} \ dy$

Let $u = lny$

$du = \dfrac{1}{y} \ dy$

$\Rightarrow I = \int \dfrac{1}{u} \ du$

$= lnu + c$

$= ln(lny)+c$

$= ln(ln(lnx)) + c$

Plugging in limits, we get 1.

Solution 202


By inspection, or letting $u=ln(ln(x))$, the answer is

$ln(ln(ln(x)))$

The definite integral is then ln(2).

I think you may have made a mistake
http://www.wolframalpha.com/input/?i=integral+of+1%2F%28xln%28x%29ln%28lnx%29%29+between+e^e+and+e^%28e^e%29

I put the brackets in the wrong place, this is the calculation I did

http://www.wolframalpha.com/input/?i=integral+of+1%2F%28xln%28x%29ln%28lnx%29%29+between+e^e+and+%28e^e%29^e

I completely understand why LOTF does it now.

what language is this
must be spanish
maybe french

Maybe one day, you'll agree on the beauty of this as well...





(by the way I am not posting this as a problem for the thread)

Just curious, to prove this would it be a **/*** problem?

It is a */**, but a difficult one.

(the joke was that it requires a mathematical argument which I love and which bananarama despises - I'll let you figure out what it is from the picture)

Maybe one day, you'll agree on the beauty of this as well...





(by the way I am not posting this as a problem for the thread)