Can't solve this difficult definite integral for the life of me

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Homework Statement


Definite integral:

[tex] \int \frac{\ln(x+1)}{x^2+1} dx [/tex] from x=0 to x=1 (sorry I don't know how to do integral boundaries with tex)




The Attempt at a Solution


I just am clueless on how to do this, I'm almost 100% sure you are not supposed to find the anti-derivative of the integral and then calculate the definite integral, but need a clever trick
to calculate the definite integral.

So the standard tricks like u-substition and partial integration do not work.


Any tips?
Thanks!
 

Answers and Replies

  • #2
LCKurtz
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It isn't an elementary integral. Maple gives an answer in terms of dilog functions and a Catalan number.
 
  • #3
HallsofIvy
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I would be inclined to use a numerical integration.
 
  • #4
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It has an algebraic solution and must be solved algebraically(so no numerical methods). I already said I thought it was not an elementary integral , and said that this one has to be solved in a clever way. Have been stuck on this one for almost a week. The answer is (pi/8)ln(2)
 
  • #5
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I would be inclined to use a numerical integration.
See my reply, sorry forgot to quote.
 
  • #6
SammyS
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Homework Statement


Definite integral:[tex] \int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx [/tex]Use _{0} and ^{1} for the limits of integration.

The Attempt at a Solution


I just am clueless on how to do this, I'm almost 100% sure you are not supposed to find the anti-derivative of the integral and then calculate the definite integral, but need a clever trick
to calculate the definite integral.

So the standard tricks like u-substitution and partial integration do not work.

Any tips?
Thanks!
Try a u-substitution that takes x=0 to u=1 and x=1 to u=0 .

Possibilities:

u = 1-t (but I think this won't work.)

[STRIKE]u = 1/(x+1) - 1 → u+1 = 1/(x+1)[/STRIKE]

[STRIKE]u = 1/(x-1) + 1 → u-1 = 1/(x-1)[/STRIKE]

Added in Edit:

(Those last two aren't as advertized. DUH !)

Try [itex]\displaystyle \ \ u=\frac{1-x}{1+x}\ .[/itex]
 
Last edited:
  • #7
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Try a u-substitution that takes x=0 to u=1 and x=1 to u=0 .

Possibilities:

u = 1-t (but I think this won't work.)

[STRIKE]u = 1/(x+1) - 1 → u+1 = 1/(x+1)[/STRIKE]

[STRIKE]u = 1/(x-1) + 1 → u-1 = 1/(x-1)[/STRIKE]

Added in Edit:

(Those last two aren't as advertized. DUH !)

Try [itex]\displaystyle \ \ u=\frac{1-x}{1+x}\ .[/itex]

I got stuck using that substitution, it just got very messy
 
  • #8
haruspex
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If you integrate by parts in the obvious way, the first term gives, intriguingly, exactly twice the desired answer. The second term is -∫atan(x)/(1+x).dx.
By a series of substitutions (x = tan(θ), θ→θ/2, θ→θ-π/4; amounting, I think, to x = tan(θ/2-π/8)) I turned that into some multiple of [itex]\int_{\pi/4}^{3 \pi/4}\frac{\left(\theta-\pi/4\right)}{sin(\theta)}.d\theta[/itex]. Splitting the range around π/2 and combining the θ term with its π-θ partner turned that into integrating cosec. Haven't checked it produces the right answer.
 
  • #9
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I notice you have 1/(x^2+1). Try a u-substitution where u= 1/(x^2+1).
 
  • #10
SammyS
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I got stuck using that substitution, it just got very messy.
For those of you helping out, as LCKurtz noted in post #2, the indefinite integral cannot be expressed in terms of elementary functions. However, the definite integral can be evaluated using a "trick".

The result of the substitution [itex]\displaystyle \ \ u=\frac{1-x}{1+x}\,,\ [/itex] is fairly clean.

[itex]\displaystyle \ \ u=\frac{1-x}{1+x}=\frac{2}{1+x}-1[/itex]

So that [itex]\displaystyle \ \ u+1=\frac{2}{1+x}\ .\ \ [/itex] This substitution behaves quite nicely.

This gives [itex]\displaystyle \ \ x+1=\frac{2}{1+u}\,,\ \ [/itex] so finding dx is easy.

Also note that [itex]\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .[/itex] Therefore, lots of things cancel out.
 
  • #11
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For those of you helping out, as LCKurtz noted in post #2, the indefinite integral cannot be expressed in terms of elementary functions. However, the definite integral can be evaluated using a "trick".

The result of the substitution [itex]\displaystyle \ \ u=\frac{1-x}{1+x}\,,\ [/itex] is fairly clean.

[itex]\displaystyle \ \ u=\frac{1-x}{1+x}=\frac{2}{1+x}-1[/itex]

So that [itex]\displaystyle \ \ u+1=\frac{2}{1+x}\ .\ \ [/itex] This substitution behaves quite nicely.

This gives [itex]\displaystyle \ \ x+1=\frac{2}{1+u}\,,\ \ [/itex] so finding dx is easy.

Also note that [itex]\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .[/itex] Therefore, lots of things cancel out.
Ok, I shall try again.

[tex] dx = \frac{-2}{(1+u)^2} du [/tex]

So the integral becomes

[tex]\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{\frac{2(1+u^2}{(1+u)^2}} \frac{-2}{(1+u)²} du = -\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{1+u^2} du = \int_{0}^{1} \frac{\ln(\frac{2}{1+u})}{1+u^2} du [/tex]

I don't see how I can get the definite integral with this substitution :-(
 
  • #12
haruspex
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Try to simplify ln(2/(1+u)), and compare it with the integral you started with.
 
  • #13
Dick
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Ok, I shall try again.

[tex] dx = \frac{-2}{(1+u)^2} du [/tex]

So the integral becomes

[tex]\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{\frac{2(1+u^2}{(1+u)^2}} \frac{-2}{(1+u)²} du = -\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{1+u^2} du = \int_{0}^{1} \frac{\ln(\frac{2}{1+u})}{1+u^2} du [/tex]

I don't see how I can get the definite integral with this substitution :-(
You are practically there. Use the rules of logs to break the integral into two parts. Notice something about one of them.
 
  • #14
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You are practically there. Use the rules of logs to break the integral into two parts. Notice something about one of them.
Oh Wow, got it now. How does one come up with this substitution?
 
  • #15
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For those of you helping out, as LCKurtz noted in post #2, the indefinite integral cannot be expressed in terms of elementary functions. However, the definite integral can be evaluated using a "trick".

The result of the substitution [itex]\displaystyle \ \ u=\frac{1-x}{1+x}\,,\ [/itex] is fairly clean.

[itex]\displaystyle \ \ u=\frac{1-x}{1+x}=\frac{2}{1+x}-1[/itex]

So that [itex]\displaystyle \ \ u+1=\frac{2}{1+x}\ .\ \ [/itex] This substitution behaves quite nicely.

This gives [itex]\displaystyle \ \ x+1=\frac{2}{1+u}\,,\ \ [/itex] so finding dx is easy.

Also note that [itex]\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .[/itex] Therefore, lots of things cancel out.
Thanks ! I got it now, how do you come up with a substitution like this?
 
  • #16
haruspex
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how do you come up with a substitution like this?
A frequent trick with nasty definite integrals is to find a way to fold the range so that the nastiness in one part cancels the nastiness in the rest. Or, what is really the same thing, reverse the whole range and reform it so that the original integral pops out as only part of the expression, and with a factor other than 1 (in this case, -1).
Since the ln(x+1) was the nasty part, that's the part to study for working out the point around which to pivot the range. To make it easier, you could first write x-1 wherever there's an x, so we get ln(x), and the range becomes 1 to 2. A natural way to pivot the range given ln(x) is to map x to 1/x, so ln(x) becomes -ln(x).
The present problem generalises to ##\int_a^b\frac{ln(x)dx}{x^2+Bx+ab}##
 
  • #17
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A frequent trick with nasty definite integrals is to find a way to fold the range so that the nastiness in one part cancels the nastiness in the rest. Or, what is really the same thing, reverse the whole range and reform it so that the original integral pops out as only part of the expression, and with a factor other than 1 (in this case, -1).
Since the ln(x+1) was the nasty part, that's the part to study for working out the point around which to pivot the range. To make it easier, you could first write x-1 wherever there's an x, so we get ln(x), and the range becomes 1 to 2. A natural way to pivot the range given ln(x) is to map x to 1/x, so ln(x) becomes -ln(x).
The present problem generalises to ##\int_a^b\frac{ln(x)dx}{x^2+Bx+ab}##
Okay, thanks!

But is there a way to see beforehand what would maybe work and what wouldn't ? or do you just have to try?
 

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