Very Hard Definite Integral

  • #1
Im studing for a math competition. One of the probles is as follows

fnInt( (Ln(x+1)/(x^2 +1)),x,0,1)

If you dont know that notation. it means Integral from 0 to 1 of Ln(x+1)/(x^2 +1)

OBVIOUSLY NO CALCULATORS... im looking for a solution, not an answer

Thanks
~rosie
 

Answers and Replies

  • #2
56
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Integration by parts

I believe that if you try an integration by parts, with u=ln(x+1) and dv=1+x^2, that should get you stated--i believe will will have to do one more integration by parts and then some long division but i think that will get you to the end
 
  • #3
HallsofIvy
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Is that
[tex]\frac{ln(x+1)}{x^2+1}[/tex]
which is what nate808 is asumming or
[tex]ln(\frac{x+1}{x^2+1})[/tex]
which is how I would read it?
 
  • #4
121
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If it's ln [tex]ln(\frac{x+1}{x^2+1})[/tex] do Integration by parts with
dv = 1 dx, and [tex]u = ln(\frac{x+1}{x^2+1})[/tex], and be prepared for alot of messy simplification and some partial fractions . Note the integral of [tex]1 /(x^2 + 1)[/tex] is [tex]tan^{-1} (x)[/tex] if it's the other integral then good luck to you.
 
  • #5
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HallsofIvy said:
Is that
[tex]\frac{ln(x+1)}{x^2+1}[/tex]
which is what nate808 is asumming or
[tex]ln(\frac{x+1}{x^2+1})[/tex]
which is how I would read it?

I would guess, based on the parentheses that he used, that it should be:

[tex]\frac{ln(x+1)}{x^2+1}[/tex], in which case parts makes the most sense. Though, I would most likely take [tex]u=ln(x+1)[/tex] and [tex]dv=\frac{dx}{x^2+1}[/tex]. Integrating dv will most likely require either partial fractions or trig. substitution, though I can't honestly say that I've worked it out.
 
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  • #6
dextercioby
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You'll probably need

[tex] \int_{0}^{1}\frac{\arctan x}{1+x}dx=\allowbreak \mbox{\func{Catalan}}-\frac{1}{2}i \ \mbox{\func{dilog}}\left( \frac{1}{2}+\frac{1}{2}i\right) +\frac{1}{2}i \ \mbox{\func{dilog}}\left( \frac{1}{2}-\frac{1}{2}i\right) [/tex]

Daniel.
 
  • #7
Hurkyl
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I don't yet see any obvious algebraic tricks. (e.g. make a clever substitution, with an easy way to relate the new integral to the old one)



My next thought is complex analysis. Maybe there's some nice contour over which you can integrate that simplifies this. My first thought is to integrate it over a rectangle, two of whose edges are [0, 1] on the real axis, and [0, H] on the complex axis, then take the limit as H goes to infinity. I haven't tried working through the algebra though, I don't know if it will be helpful.

Note that if you try this approach, you either have to use a small semicircle to avoid the pole at i... or use partial fractions to break the integrand up into two parts, and do an upwards strip for one part, and a downwards strip for the other.



My third thought is to try a power series: you can find the power series for the integrand, and you can integrate that power series -- maybe you can recognize the resulting infinite sum.



Maybe I'll think about this a bit at work and have a better idea for you later.
 
  • #8
Prolly should read ln(x) in the numerator. Then you have int ln(x)/(x^2+1),0,1

Widder shows how this integral is equal to -(pi^2)/8.

1) Substitute 1 - x^2 + x^4 - x^6... for the 1/(1-x^2) term

2) Interchange summation and integration

3) integrate (x^2k)*ln(x+1) by parts

4) Sum the infinite series
 
  • #9
The value of your original (with derivation) appears in Todhunter's Integral Calculus Text (circa 1860's).... it is available via www.elibron.com

lots of ingenious integration devices in there actually...:bugeye:
 
  • #10
factor x^2 + 1 = (x+i)(x-i) and use partial fraction A/x+i B/x-i
 
  • #11
dextercioby
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The value of your original (with derivation) appears in Todhunter's Integral Calculus Text (circa 1860's).... it is available via www.elibron.com
lots of ingenious integration devices in there actually...:bugeye:

The book isn't free unfortunately.
 
  • #12
Curious3141
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This is a problem from the 2005 Putnam competition (problem A5, to be precise). It can be solved with a choice of nifty substitutions :

Problems : http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005.pdf [Broken]

Solutions : http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005s.pdf [Broken]
 
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  • #13
1
0
Integral

Integral from 0 to 1 of ln(x+1)/(x^2 +1) dx isn't that hard actually but I only know one way to do it.
1/(x^2+1) = (arctg x)' ... (ie the derivate of arctan x) tg(x)=tan(x)
you change the variable and the integral becomes
integral from 0 to pi/4 of ln(1+tg(x)) dx
by making the following substitution you should get it done:
x=pi/4-u
dx=-du
the integral becomes
- integral from pi/4 to 0 of ln(1+tg(pi/4-u)) du =
integral from 0 to pi/4 of ln(1+((1-tg(u)/(1+tg(u))) du =
integral from 0 to pi/4 of ln(2/(1+tg(u)) du =
integral from 0 to pi/4 of ln2 du - integral from 0 to pi/4 of ln(1+tgu) du
and you're back from where you started.
That means our integral I is
I=pi/4*ln2-I => 2I=pi/4*ln2 => I=pi/8*ln2
 

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