Incredibly Hard Integration..or not... (1 Viewer)

[lucifel]

This integration has been at me for weeks, and I still have NO IDEA how to approach it, help would be appreciated please.


∫ln (ln x).dx

 

[currysauce]

ln²x²/2 + c?

 

[ephemeral]

yeah, that looks right.

 

[Slidey]

ln(lnx)=1.ln(ln(x))
Integral ln(lnx)
=xln(lnx) - integral 1/ln(x)

Hmm.

integral x(ln(lnx))/x dx
u=ln(x)
du=1/x
integral e^u.lnu du
=e^u.lnu - integral (e^u)/u du
=lnx.ln(lnx) - integral x/lnx dx

Hmm.

 

[lucifel]

it doesn't work does it Slide Rule? and for the (ln^2 x^2)/2 answer, i don't think that works.

 

[Slidey]

Where did you get it from? If it wasn't a past paper or 4u textbook, then you do realise it probably requires all sorts of fancy stuff, such as PolyLogs?

 

[nit]

where did you come across this question?

Edit: Slide, you pre-empted me there.

 

[Will Hunting]

Use the substitution, ln x = y, x = e^y, dx = e^ydy

Now,

∫ln(ln x)dx = ∫e^y(ln y)dy

IBP, where,

u = ln y, u' = 1/y
v' = e^y, v = e^y

I = e^y(ln y) - ∫e^ydy/y

Again,

∫e^ydy/y = ∫e^ydy/ln e^y = e^y/y + ∫e^ydy/y²

*Yawwwwn*... I'm sleepy. Soz about the incompleteness. I'll give it another bash tomoz.

 

[ngai]

This integration has been at me for weeks, and I still have NO IDEA how to approach it, help would be appreciated please.


∫ln (ln x).dx

maple says its ln(ln(x))*x+Ei(1,-ln(x))
whatever Ei is....

 

[haboozin]

maple says its ln(ln(x))*x+Ei(1,-ln(x))
whatever Ei is....

yea, this isn't a 4u integration so it should be in the beauty and elegance section..

 

[martin]

The integrator http://integrals.wolfram.com/ (which uses mathematica) gives

integral(Log[Log[x]]) = xLog[Log[x] - LogIntegral[x]

A bit of research shows that LogIntegral[x] = integral from 0 to x of dt/logt. So it seems that it isn't possible to do in terms of elementary functions. This is true of most random integrals that you choose.

 

[Slidey]

^ Indeed. That's the basic result you get when you use integration by parts once.

ln(ln(x))=1.ln(ln(x))
u=ln(ln(x)), v=x
u'=1/(xln(x)), v'=1
Integral ln(ln(x)) = uv'
=uv - integral u'v
=xln(ln(x)) - integral 1/ln(x)

 

[lucifel]

No I didnt' think it was a 4U integration, oh well, i just saw an easier one where the there was an extra 'x' in the denominator, so i thought what would happen if i took that out, obviously made the integration that much more difficult.

Oh well, still a solution would be appreciated, even though I may not understand it.

 

[Slidey]

You don't understand; there probably isn't a solution.

If there is, it's beyond the capability of even Mathematica.

 

[m_isk]

substituting u=lnx, you get integrate e^u lnu...can this be integrated? :rofl: (i'm thinking parts)

 

[lucifel]

i severly doubt there is no solution per se, there could be one, just way beyond the scope of 4U, it doesn't mean there is no solution to it.

 

[m_isk]

i severly doubt there is no solution per se, there could be one, just way beyond the scope of 4U, it doesn't mean there is no solution to it.

why?? it could be IMPOSSIBLE...eg solve lnx=0 No solution!!

 

[lucifel]

thats undefined more than no solution, integrations can't be 'undefined' as such.

 

[m_isk]

hhmmmm..good point, but still, if i siad solve for x: lnx=0, there is no solution why???because ln0 is undefined..anyway this is irrelevant does anybody out there have a solution!?!?!?!

 

[SeDaTeD]

The solution has already been given, just read up a few posts. However the solution cannot be understood with elementary calculus.