• Support PF! Buy your school textbooks, materials and every day products Here!

Tough Integral

  • Thread starter awvvu
  • Start date
188
1
This integral came up while trying to find the potential of a uniformly charged rectangle.

[tex]\int \log(\sqrt{a^2+x^2} + b) dx[/tex]

Integrator gives a pretty long expression involving inverse tangents so I'm not sure where to begin at all. I tried integrating by parts once, taking u to be the whole expression, but it just makes it messier. I also tried the trig subtitution:

[tex]x = a \tan(\theta)[/tex]
[tex]\int a \log(a \sec(\theta) + b) \sec^2(\theta) d \theta[/tex]

But that's not any easier to integrate.
 
Last edited:

Answers and Replies

1,750
1
Damn nobody? I am stumped as well, I have this problem at the back of my head ... hopefully something pops up :)
 
492
1
I don't know if this is going to work but how about:

assuming the log is of base e

[tex]y=\int \log(\sqrt{a^2+x^2} + b) dx[/tex]
[tex]e^y=\int (\sqrt{a^2+x^2}+b) dx[/tex]

solve the RHS and then take the ln of the resulting solution
 
Dick
Science Advisor
Homework Helper
26,258
618
I don't know if this is going to work but how about:

assuming the log is of base e

[tex]y=\int \log(\sqrt{a^2+x^2} + b) dx[/tex]
[tex]e^y=\int (\sqrt{a^2+x^2}+b) dx[/tex]

solve the RHS and then take the ln of the resulting solution
I don't know how to do the integral easily, but I know you DEFINITELY can't do that. If this is related to the potential of a uniformly charged rectangle I'd suggest you post the steps you used to get to that integral. Shouldn't it be a double integral? There may be an easier approach that doesn't involve evaluating that integral.
 
1,631
4
This integral came up while trying to find the potential of a uniformly charged rectangle.

[tex]\int \log(\sqrt{a^2+x^2} + b) dx[/tex]

Integrator gives a pretty long expression involving inverse tangents so I'm not sure where to begin at all. I tried integrating by parts once, taking u to be the whole expression, but it just makes it messier. I also tried the trig subtitution:

[tex]x = a \tan(\theta)[/tex]
[tex]\int a \log(a \sec(\theta) + b) \sec^2(\theta) d \theta[/tex]

But that's not any easier to integrate.
have you tried using integration by parts? I think it would simplify thigs a lot. And after that probbably a trig substitution would work, i am not sure though, i haven't tried it myself.
 
188
1
If this is related to the potential of a uniformly charged rectangle I'd suggest you post the steps you used to get to that integral. Shouldn't it be a double integral? There may be an easier approach that doesn't involve evaluating that integral.
Okay, there's a rectangle with uniform charge density [itex]\delta[/itex] with one corner at the origin and the other corner at (L, H). And we want to find potential as a function of (x, y), so x and y are constants in the following integral.(this is gonna be fun to type up).

[tex]V = K \int_R \frac{dQ}{r_i}[/tex], where [tex]r_i = \sqrt{(x - x_i)^2 + (y - y_i)^2}[/tex] and [tex]dQ = \delta dA = \delta dx_i dy_i[/tex]
[tex]K \delta \int_0^H \int_0^L \frac{1}{\sqrt{(x - x_i)^2 + (y - y_i)^2}} dx_i dy_i[/tex]

Now doing the substitution [tex]u = x - x_i[/tex] and [tex]v = y - y_i[/tex]:

[tex]K \delta \int_{y - H}^{y} \int_{x - L}^{x} \frac{1}{\sqrt{u^2 + v^2}} du dv[/tex]

Then doing the trig substitution [tex]u = v \tan(\theta)[/tex], so [tex]du = v \sec^2(\theta) d\theta[/tex]:

[tex]K \delta \int_{y-H}^{y}\int \frac{v \sec^2(\theta)}{\sqrt{v^2 \tan^2(\theta) + v^2}}d\theta dv[/tex]
[tex]K \delta \int_{y-H}^{y}\int \sec(\theta) d\theta dv[/tex]
[tex]K \delta \int_{y-H}^{y} \left[ \log(\sec(\theta) + \tan(\theta)) \right] dv[/tex]
[tex]K \delta \int_{y-H}^{y} \left[ \log(\sqrt{u^2 + v^2} + u) - \log(v) \right]_{u = x - L}^{u = x} dv[/tex]
[tex]K \delta \int_{y-H}^{y} \log(\sqrt{x^2 + v^2} + x) - \log(\sqrt{(x - L)^2 + v^2} + x - L) dv[/tex]

The last integrals are of form: [itex]\int \log(\sqrt{a^2+x^2} + b) dx[/itex] like above. I think the left integral is much simpler because a = b in that case, and stuff cancels out. But the right one is a lot more complicated.

edit: oh wait, a = b for the right integral too, so that makes it a lot easier!

[tex]\int \log(\sqrt{a^2 + x^2} + a) = x \log(\sqrt{a^2 + x^2} + a) + a \log(\sqrt{a^2 +x^2} + x) - x[/tex] (from Integrator)

I still don't know how to do this integral though.
 
Last edited:

Related Threads for: Tough Integral

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
1K
Top