Improper integrals with singularities at both endpoints.

[Lavabug]

Homework Statement

Study the continuity of $$\int\frac{dx}{x \sin x}$$ from 0 to pi/2
That's 1/(xsinx), latex isn't showing up clearly for me.

I've been having a go at simply solving it as an indefinite integral to evaluate it but I keep ending up with more complicated expressions if I try integrating by parts.

 

[micromass]

The integral $$\int{\frac{1}{x*\sin(x)}dx}$$ is a notorious example of an integral which can't be solved using elementary functions.
So the only thing you can do is showing that the definite integral converges by using tests like the comparison test and stuff. You can probably find the solution to the definite integral, but that needs a lot of advanced math...

 

[HallsofIvy]

And, by the way, there is no singularity at $\pi/2$.

 

[jackmell]

Homework Statement

Study the continuity of $$\int\frac{dx}{xsenx}$$ from 0 to pi/2
That's 1/(xsinx), latex isn't showing up clearly for me.

I've been having a go at simply solving it as an indefinite integral to evaluate it but I keep ending up with more complicated expressions if I try integrating by parts.

I think you should more clearly define the problem. You're asking about the continuity of the function:

$$f(u)=\int_0^{u}\frac{1}{x\sin(x)}dx,\quad u\geq 0$$

Well surely if the lower limit wasn't zero, the function would be continuous in the interval $0+\epsilon<u<\pi$

so perhaps you could focus on analyzing it's continuity at zero. Does the integral:

$$\int_{0}^{1} \frac{1}{x\sin(x)}dx$$

converge? If it does, then the function f(u) would be continuous in that interval. If it doesn't, can you prove why it does not converge?

 

[Lavabug]

Thanks hallsofivy, I missed that.

I've been having a look at the comparison test method here:
http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

So basically, since my problem is at the lower limit of integration(0), I need to figure out the convergence using by taking the limit of the expression as t approaches 0(from the right?). I'm still facing the problem of solving the integral.

The site I posted says I need to test for a "larger function" and see if it converges. I'm not quire sure I'm interpreting this method correctly. Is the larger function in this case 1/x? 1/sinx? Both seem to diverge as x approaches 0.

 

[micromass]

Well, if you want to do the comparison test, then there are two approaches

1) you can find a larger function that converges, then your original function converges
2) you can find a smaller function that diverges, then your original function diverges.

Now, what can you tell if you compare with 1/x?

 

[Lavabug]

First of all, is 1/x smaller than my given integral? (I think so, how can I know for sure?) It doesn't converge, so neither does my given integral? How do I prove this formally?

 

[micromass]

Yes, that is correct. Try to formalize this...

 

[Lavabug]

Yes, that is correct. Try to formalize this...

Thanks, here's what I've got, would appreciate any comment:

[PLAIN]http://img831.imageshack.us/img831/7277/p10008950.jpg

 

[micromass]

Yes, that seems to be correct. But there's a small typo in there. You wrote

$$\lim_{t\rightarrow 0+}{[ln(t)-ln(\pi/2)]}$$

But shouldn't it be $$ln(\pi/2)-ln(t)$$ instead?

 

[Lavabug]

whoops yeah, so -ln t would become positive infinity?

 

[micromass]

Correct! It seems like you've got it

 

[Lavabug]

Thanks a lot! ;)

 

[Lavabug]

Question about the comparison method: I'm not too quick on telling if the 1/x^p function you choose to compare an integral to is "bigger" or "smaller", what's a surefire way of telling? Just plugging in any number and evaluating?

 

[micromass]

No, plugging in an answer is not sufficient.

You'll need to show that

$$\frac{1}{x}\leq \frac{x\sin(x)}$$

which is equivalent to (since x is positive):

$$x\sin(x)\leq x$$

You'll need to show that this holds for every x...

 

[Lavabug]

Not sure I understood. As a general rule, should I pick a 1/x^p function with a denominator I can easily tell is bigger/smaller than the denominator in the integral for any x? If x^p is smaller, that implies 1/x^p jumps to infinity faster than the other function, therefore is "bigger"?

 

[micromass]

Yes, that reasoning is correct.

 

[Lavabug]

Thanks a lot for the quick replies!