Using computer to solve integral

[lisawoods]

I have attachted a word document with the integral on. Does anyone have a suitable suggestion on how to solve this?

 

[Pseudo Statistic]

Two things.
1) Put this in the homework section, PLEASE!
2) For some reason your doc won't open with AbiWord. Care to paste it here?

 

[bomba923]

Why not just use \LaTeX ?
~

 

[dextercioby]

Your integral is ugly enough not to be solved analytically. Since you're asked for an expression depending on "l" (and not for the antiderivative of the integrand), you could use a computer to see what it gives you.

Daniel.

 

[lisawoods]

Integral progress

I have actually progressed with it now. The whole point was not to use a computer to solve it but to do it by hand.

I am now stuck on this bit any ideas? oh the integral is with respect to dz

 

[benorin]

That's not pretty

\int \frac{\cos(2z)}{1+2z} dz = \frac{1}{2}\mbox{Si}(1+2z)\sin 1+\frac{1}{2}\mbox{Ci}(1+2z)\cos 1

where the functions Si(x) and Ci(x) are the Sine Integral and Cosine Integral functions (these are not elementary functions), respectively.

Alternatively, look here (Sine Integral) and here (Cosine Integral).

 

[quantumdude]

I have actually progressed with it now. The whole point was not to use a computer to solve it but to do it by hand.
I am now stuck on this bit any ideas? oh the integral is with respect to dz

Lisa,

It might expedite matters if you typed out your problems using LaTeX. Click on the image below to see the code that was used to generate it (won't work if you have a pop-up blocker on).

\int\frac{\cos(2z)}{1+2z}dz

Now I have two questions for you:

1.) Is that integral supposed to be definite, or indefinite?

2.) If it is definite, then are you familiar with the theory of residues? If it is indefinite, then I think you are doomed to either look it up in a table or use mathematical software.

 

[lisawoods]

Integral

its a definite integral with limits from 0 to L/2 where L is greater than 0.

How would you suggest I proceed with solving it using residues

Lisa

 

[quantumdude]

Arrrgh...I was hoping that it went from -\infty to \infty because then a contour integral would be pretty straightforward, as your integrand only has one simple pole at z=-\frac{1}{2}.

I'll have to hit the books and review how to handle this. In the mean time, follow the lead given by benorin.

 

[lisawoods]

Integral

Tom, did you get to talk to your math tutors?