Integration

[twentysix by twentyfive]

Hi all you bright things. Can you help on this one please?

Integrate Ln[x]/(1+x^2) dx

Or suggest a good substitution which splits the integral in such a way that I can show the answer is zero when the limits are zero to Infinity.

Substitutions I get then get stuck with are x=e^u, x=Tan, ....

I can do this easily for Ln[x]/(1+x)^2 dx (it can be done by parts but a nice answer falls out when not done by parts).

 

[marinyork]

Erm, quick thoughts:-
ICBST ∫arctan(x).dx= xarctan(x) - (1/2) ln (1+x[sup]2[/sup]) + C.

 

[upsidedown]

I think i'd be tempted to just round it up.

 

[marinyork]

D'oh. An easier substitution is z=x. Residue theorem. Integrate around the simple pole z=i. If you can't do it I'll write something up later.

Just use the principal logarithm value ln(i) = πi/2. Use some contour that makes sense and doesn't muck it up. Set all bits of this equal to the value from the residue theorem. Strip out whichever bit you want. Take real/imaginary parts as as radius → ∞. As there is no real part in whatever it is you have constructed then the answer is obviously zero!

 

[MrGrumpy]

fook i used to be able to do these in my sleep

about 23yrs ago

In fact I`ve just been reaquanting myself with trigonometry since my wife started her Uni course!


(d ((Log[1 - I x] + Log[1 + I x]) Log[x] + PolyLog[2, -I x] + PolyLog[2, I x])) / 2

 

[MrGrumpy]

http://www.wolframalpha.com/input/?i=Ln[x]/(1+x^2)+dx

 

[Aperitif]

Looks like a crash diagram between iLB and MyCyclingLogs to me...phew!

 

[marinyork]

(d ((Log[1 - I x] + Log[1 + I x]) Log[x] + PolyLog[2, -I x] + PolyLog[2, I x])) / 2

I've been thinking about this and am far from convinced it will work. Could you show your working as to why this works for -ix and ix?

 

[MrGrumpy]

oh look whats that over their --------------------------------------------------------------------------------------------------------------------------------------->
















does a runner.....

 

[marinyork]

Oh it's typing things into Mathematica is it . Seems a good way of doing it - if you can get the function to behave and I'm not totally convinced of that. I still think using elementary complex analysis is a better way of doing it than monkeying around with analytic continuation and other special functions though or even some convergence test. Seems an elegant way of sorting out a problem from real analysis with complex analysis.

 

[darth vadar]

I won't be able to sleep now tonight worrying about this.

 

[vernon]

It's barely keeping me awake

 

[marinyork]

It's barely keeping me awake

You are actually asleep and dreaming about this thread (which isn't real).

 

[darth vadar]

Sounds like the kind of stuff that would be discussed at the world's most boring dinner party.

 

[marinyork]

Sounds like the kind of stuff that would be discussed at the world's most boring dinner party.

What sort of things are you interested in?