Improper integrals

[nasillmatic20]

I seem to be stuck on the mechanics of this problem:
∫(dx/1+x^2) (on interval -∞ to ∞)
which I separated into the addition of the two separate integrals with 0 as the midpoint. I just can't remember how to integrate the function 1/(1+x^2)... is it just simply ln|1+x^2| ?

 

[ChaoticLlama]

take the derivative of y = ln|1 + x²| to find that y' = 2x / (1 + x²). Not the correct answer.

Think of your derivatives of trigonometic inverse functions (ie arcsin(x), arcos(x), arctan(x))

Reply back if you get stuck again.

 

[nasillmatic20]

alright, so I figured that the integral is arctan(x)... but I can't compute the limit of x as it approaches infinity or negative infinity... on my graphing calculator I can't get a value higher/lower than +/- 1.57, but there must be a better way of conducting this process?

 

[soroban]

Hello, nasillmatic20!

\(\displaystyle \L\int^{\;\;\;\infty}_{-\infty}\frac{dx}{1\,+\,x^2}\)

And you got:\(\displaystyle \;\arctan(x)\,\bigg]^{\infty}_{-\infty}\) . . . good!

Recall the meaning of arctangent . . . it's an angle.
\(\displaystyle \;\;\theta \,=\,\arctan x\;\;\Rightarrow\;\;\tan\theta\,=\,x\)


For the upper limit, we have: \(\displaystyle \,\tan\theta\,=\,\infty\)
\(\displaystyle \;\;\)What angle has a tangent of infinity? . . . Answer: \(\displaystyle \,\frac{\pi}{2}\)

For the lower limit, we have: \(\displaystyle \,\tan\theta\,=\,-\infty\)
\(\displaystyle \;\;\)What angle has a tangent of negative infinity? . . . Answer: \(\displaystyle \,-\frac{\pi}{2}\)

Got it?

 

[nasillmatic20]

Yes, thanks so much for all the help.... I do have another similar problem that seems to be eluding me however.
∫(dx/ root(9-x^2)) on interval from 0 to 3
so I know that it is discontinuous at 3, so I set the limit t->3- but I'm not sure how to assess the integral.


I also worked through the problem:
∫(dx/x) from -1 to 1
and assessed the side limits of 0 for the integral, and found that neither of the limits existed, so the improper integral is divergent... just wanted to check the validity of this answer.

Thanks!

 

[royhaas]

Use the substitution \(\displaystyle x=3sin(u)\) to evaluate the integral in terms of the arcsine function.

 

[nasillmatic20]

Ok, I understand that arcsin(x) = ∫ dx/root(1-x^2) from 0 to x but I'm not sure how to apply that and the identity that you gave me to this problem.... could anyone help clarify?

 

[Gene]

x=3sin(u)
dx=3cos(u)du
dx/sqrt(9-x²) =
3cos(u)du/sqrt(9-9sin^2(u)) =
3cos(u)du/3cos(u) =
du
The integral = u
x=0 => u=0
x=3 => u=pi/2
pi/2-0=pi/2