The hard integral thread.

@Zacken @SeanFM @Kummer @atsruser

Showing all your working, demonstrate that

$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx = 4 \pi \, \text{arccot}{\sqrt{\phi}}$

In case anybody wants to try this: http://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1/565626#565626

@Zacken @SeanFM @Kummer @atsruser

Showing all your working, demonstrate that

$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx = 4 \pi \, \text{arccot}{\sqrt{\phi}}$

This is an easy one requiring no more than A level techniques.

Show that:

$\displaystyle \int_0^1 (^ne)^x + xe^{x +e^x + (^2e)^x + \cdots + (^{(n-1)}e)^x} \ dx = ^{(n+1)}e$

where $^k e$ is the k'th tetration of $e$ (e.g. $^3 e = e^{e^e}$)

@Zacken @SeanFM @Kummer @atsruser

Showing all your working, demonstrate that

$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx = 4 \pi \, \text{arccot}{\sqrt{\phi}}$

Whats this. Ur homework.


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@Zacken @SeanFM @Kummer @atsruser

Showing all your working, demonstrate that

$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx = 4 \pi \, \text{arccot}{\sqrt{\phi}}$

Is the answer correct? I get the answer to be $^{(n)}e$?

I'm too busy to try this at the moment, but it looks like it may succumb to a $x \to \frac{1}{x}$ substitution.

No, it's wrong. I misread my result. I'll edit.

In that case,

First observe that $\frac{d}{dx}\left((^ne)^x\right) = e^{x +e^x + (^2e)^x + \cdots + (^{(n-1)}e)^x}$. Then
$\displaystyle \int_0^1 (^ne)^x + xe^{x +e^x + (^2e)^x + \cdots + (^{(n-1)}e)^x} \ dx = \underbrace{ \int_0^1 (^ne)^x\ \mathrm{d}x}_{(1)} + \underbrace{\int_{0}^{1} xe^{x +e^x + (^2e)^x + \cdots + (^{(n-1)}e)^x}\ \mathrm{d}x}_{(2)}$

Now consider (2). Use I.B.P (differentiate $x$ and integrate the exponential term), then

$\displaystyle \int_0^1 (^ne)^x\ \mathrm{d}x + \big[x (^ne)^x\big]_{0}^{1} - \int_0^1 (^ne)^x\ \mathrm{d}x = \big[x (^ne)^x\big]_{0}^{1} =\ ^ne$ as required.

I agree with this but wouldnt it be easier just to note the use of the product rule in the first integral? The answer then follows straight away

I'm not sure what you mean by "use of the product rule" here, but, if you are familiar with the derivative of the tetrated term, then the result follows in one line by IBP on the first term:

$\displaystyle \int_0^1 (^ne)^x \ dx = [x(^ne)^x]_0^1 - \int_0^1 x \frac{d }{dx} (^ne)^x \ dx$

I agree with this but wouldnt it be easier just to note the use of the product rule in the first integral? The answer then follows straight away

Evaluate $\displaystyle \int^{\frac{\pi}{2}}_{0} \arctan{\left(\cos^2{x}\right)}\ \mathrm{d}x$

If $\displaystyle I(\alpha) = \int_0^{\pi/2} \arctan (\alpha \cos^2 x) \, \mathrm{d}x$ then:



is about as far as I can get. I was considering a $t \mapsto \frac{1}{t}$ sub but that doesn't seem to get it to yield.

As in $[br]\dfrac{d}{dx}(f(x)g(x)) = f'(x)g(x)+g(x)f'(x)$ ? so just observing that if we let $f(x) = x ,g(x) = (^ne)^x$ we get the integral in the question i.e. we can just eval. $x (^ne)^x$ between 0 and 1.

lol sorry if being really dumb or something XD

Right. I see what you mean - that is essentially equivalent to the IBP approach that I suggested, so it's fine. The real trick to this problem is to recognise the derivative - there's nothing else to it, really.