Expert: Paul Klarreich - 9/21/2009

Question
QUESTION: I have tried IBP, substitutions, and z substitutions
and I just can not figure this one out. I found
this problem on-line. I know that it is improper integral
but can't solve the integration.
Any tips or Hints?
Anything would be appreciated.

ANSWER: Questioner: Mark
Country: United States
Category: Calculus
Private: No
Subject: Integration
Question:

I have tried IBP, substitutions, and z substitutions
and I just can not figure this one out. I found
this problem on-line.

>> Where did you find it?  Perhaps I can look.

I know that it is an improper integral
but can't solve the integration.
Any tips or Hints?
Anything would be appreciated.  
.....................................
I don't think there is a 'closed form' integral for this.  My usual method for this kind of stuff is rationalizing substitutions:

{  sin  x
| --------- dx
}  sqrt(x)

u = sqrt(x)

u^2 = x

2udu = dx

{  sin(u^2)
| --------- 2u du
}     u

{
| 2 sin(u^2) du
}

But this is not integrable either.

HOWEVER, you did say this was an improper integral.  If all you have to do is tell whether it converges or diverges, maybe we can do that without integrating.

P.S. You have to write bigger if you send me a scan of your work.  My eyes ain't what they used to be.


---------- FOLLOW-UP ----------

QUESTION: Thank you.
Yes, I was asked to prove if it converged or diverged.
Do I just take limits at infinity. I belive the squeeze thorem
would allow to do it. Have to go back a few chapters though.

Thanks again,
Mark

Answer
I don't think you need the squeeze theorem, but you do need some stuff about infinite series.  (Usually it is in the same chapter with improper integrals.)

I assume you want:

{ inf
|    sin x / sqrt(x)  dx
} 2pi

Suppose we write that as:

SUM (k = 2 to infinity)

{ (k+1)pi
|    sin x / sqrt(x)  dx
} k pi

Now

{ (k+1)pi
|    sin x dx = plus-or-minus 2, depending on whether k is even or odd.
} k pi

Now if k is even, we have (this part has sin x above the x-axis):

{ (k+1)pi
|    sin x / sqrt(x)  dx
} k pi

<=

{ (k+1)pi
|    sin x / sqrt(k pi)  dx
} k pi

= 2/sqrt(k pi)

And if k is odd,

= - 2/sqrt(k pi)

So you have this summation:

SUM[k = 2 to inf] 2(-1)^n/sqrt(k pi)

Now I think the rule is that an alternating series (+,-,+,-,+,-,+,-...) will converge if the absolute values approach zero monotonically, which they do in this case.