Integration question

  • #1

Homework Statement



[tex]\int\frac{dx}{(R^{2}+x^{2})^{3/2}}[/tex]

Homework Equations




The Attempt at a Solution



[tex]\mbox{Let }x = R\\tan\theta \Rightarrow dx = R\\sec^{2}d\theta[/tex]
[tex]x^{2} = R^{2}\\tan^{2}\theta[/tex]
[tex]\Rightarrow R^{2} + x^{2} = R^{2}\\( 1 + tan^{2}\theta)[/tex]
[tex]\Rightarrow R^{2} + x^{2} = R^{2}\\sec^{2}[/tex]
[tex]\Rightarrow (R^{2} + x^{2})^{3/2} = R^{3}\\sec^{3}\theta[/tex]

[tex]\mbox{Therefore, }\int\frac{dx}{(R^{2}+x^{2})^{3/2}}[/tex]
[tex]=\int\frac{R\\sec^{2}\theta}{R^{3}\\sec^{3}\theta}\\d\theta[/tex]
[tex]=\frac{1}{R^{2}}\\ \int^{\infty}_{-\infty}cos\theta\\ d\theta[/tex]
[tex]=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty} \int^{N}_{-N}cos\theta\\ d\theta[/tex]
[tex]=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin\theta)\mid^{N}_{-N}[/tex]
[tex]=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(mbox{sin N + sin N})[/tex]
[tex]=\frac{2}{R}\stackrel{lim}{n\rightarrow\infty}mbox{sin N}[/tex]

and then??????????????????????
 

Answers and Replies

  • #2
799
0
It's simply that [tex]lim_{x\rightarrow \infty} (sin x)[/tex] does NOT exist.
 
  • #3
But the integral has an answer, and it's 1!
 
  • #4
799
0
But the integral has an answer, and it's 1!
The last lim you got doesn't exist. The wrong step here is when you changed the variable from x to theta. I guess the upper and lower limits of x are [tex]\infty[/tex] and [tex]-\infty[/tex], right? The relation between x and theta is: [tex]x=Rtan\theta[/tex]. Since tan is a periodic fuction, tan(theta) cannot be defined if theta=infinity. Thus, you must choose another upper and lower limits for theta.
Hint: there is an infinite number of values for theta so that tan(theta) = infinity.

If you get the right limits, you will find the integral's value is 2 (not 1).
 
  • #5
34,311
5,949
The last lim you got doesn't exist. The wrong step here is when you changed the variable from x to theta. I guess the upper and lower limits of x are [tex]\infty[/tex] and [tex]-\infty[/tex], right? The relation between x and theta is: [tex]x=Rtan\theta[/tex]. Since tan is a periodic fuction, tan(theta) cannot be defined if theta=infinity.
You're on the right track here, but your argument has a flaw. The OP can undo the substitution and use the original limits of integration, or can find the new limits of integration as you are suggesting. The relationship is x = Rtan(theta). If x = infinity, that implies that Rtan(theta) = infinity, not that theta = infinity. x --> infinity if theta --> pi/2 from the left.
Thus, you must choose another upper and lower limits for theta.
Hint: there is an infinite number of values for theta so that tan(theta) = infinity.

If you get the right limits, you will find the integral's value is 2 (not 1).
To the OP:
Since the integrand you ended with is cos(theta), an even function, use the symmetry of the function like this:

[tex]\int_{-b}^b f(x) dx = 2 \int_0^b f(x) dx[/tex]
 
  • #6
Char. Limit
Gold Member
1,204
14
But there's still the problem that you're using an improper integral where you should be using an indefinite integral. There shouldn't be any bounds that you're integrating on... You can't just change indefinite to definite like that.
 

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