Strategies for Solving Complex Integrals with Polynomial Functions

[pivoxa15]

<br /> \int_{0}^{a}\frac{1}{1+y(x)^2}sin(z(x))dx<br />

where y(x) and z(x) are polynomials of degree one i.e ax+b. a,b constants.

I have tried integration by parts but that leads to ever more complex functions to integrate so that doesn't work.

If we restrict the magnitude of the function y(x) and expand y(x) in a geometric series, that dosen't seem to work either because the sine function dosen't reduce and integration or differentiating the geometric series get new results each time. Any help would be appreciated.

 

[Mindscrape]

I think after a couple integration by parts it should work out. If not then write back.

 

[Benny]

Please do not make such misleading comments. I mean, at least think about the problem before suggesting that something decent will come out of the working. I don't mean to sound harsh but misleading 'advice' is much worse than no advice at all. It won't "work out" unless you like dealing with functions/expressions which are very difficult to work with on paper.

As for the question, depending on what sort of problem you're working on, you don't always need to evaluate the most obvious thing (in this case the integral you posted).

 

[pivoxa15]

It won't "work out" unless you like dealing with functions/expressions which are very difficult to work with on paper.

What do you mean? Are you suggesting another method apart from integration by parts? That method seems to be generating ever more complex function which seem impossible to "work out", ever.

Depending on what sort of problem you're working on, you don't always need to evaluate the integral.

You are doing the same assigment as me aren't you? I am looking at q1 b) iii. I see a method to do it without integrating, which is to use the geometric series and than get upper and lower limits of that series. But that would introduce an additional assumption that epsilon must be small instead of being an arbitary number greater than 0.

Or I could just make the boundary conditions between 0 and 1 but that seems to trivial. What is (2x-L)^2 for if that is the case?

 

[pivoxa15]

I think after a couple integration by parts it should work out. If not then write back.

As mentioned, the integrals get more complicated. i.e. you get back the original integrals plus a polynomial attached which is not good.

 

[Benny]

I don't have the solution so I can't guarantee that the following is correct.

1. Yes, definitely avoid evaluating the integral that you posted.

2. At this stage it isn't a matter of whether or not it's trivial but rather a matter of whether or not you can actually get a bound out of that.

As I said before, I wouldn't know if I'm correct but I managed to obtain bounds which required no restriction on epsilon other than that which was given - that it needs to be positive.

 

[pivoxa15]

There seems to be a typo for q2 a) with regards to T<t<T. What should it be?

 

[Benny]

I took it as meaning -T < t < T.

Edit: BTW my first paragraph wasn't directed towards you.

 

[pivoxa15]

I don't have the solution so I can't guarantee that the following is correct.

1. Yes, definitely avoid evaluating the integral that you posted.

2. At this stage it isn't a matter of whether or not it's trivial but rather a matter of whether or not you can actually get a bound out of that.

As I said before, I wouldn't know if I'm correct but I managed to obtain bounds which required no restriction on epsilon other than that which was given - that it needs to be positive.

With your point 2, does that mean you think my statement "make the boundary conditions between 0 and 1" is correct?

When considering large epsilon, Uy(x,y) should be 0 everywhere except at x=L/2, which would be 1. But the Fourier series cannot pick that up can it so the boundary conditions are still 0 and 1, which would make the final solutions differ by the absolute value of L(x,y) (where L(x,y) satisfies the Laplace's equation and inhomogenous BC).

Going with this method further, I obtained a bound for the change (del) in potential (which I take to be the general solution U(x,y)) to be 0<delU(x,y)<L/2 It seems like a neat solution which is usually a good indication of success. But I didn't use (2x-L)^2 which is a worry. I don't think the L/2 I got for the upper bound is related to the L/2 I mentioned earlier or is it?

 

[Benny]

There might be more than one acceptable answer but I used different reasoning to you so whether or not I think your comment about the boundaries is correct isn't really going to be helpful.

 

[matt grime]

Obviously ont everyone here has the sheet you're discussing, so most of that is going to be lost on us.

As for the original question, you can assume that y(x)=x (if it is not a constant), after a change of variable, and you may of course use compound angle formulae. Dually, you can assume that z(x)=x instead if that helps.

I don't know how much that helps.

 

[pivoxa15]

Obviously ont everyone here has the sheet you're discussing, so most of that is going to be lost on us.

As for the original question, you can assume that y(x)=x (if it is not a constant), after a change of variable, and you may of course use compound angle formulae. Dually, you can assume that z(x)=x instead if that helps.

I don't know how much that helps.

Is it a problem if in my specific situtation, y(x)=ax+b where a and b are nonzero constants but z(x)=cx where c is a nonzero constant.

 

[jpr0]

You can get the answer in terms of the exponential integral function, but your resulting expression is very ugly. Maybe there is an easier way but I can't see it.
Let me know if you'd like to see the exponential integral function represantion.

 

[jpr0]

Writing the integral as

<br /> \int_{0}^{s}dx\frac{\sin q\left(x\right)}{1+p^{2}\left(x\right)}<br />

where q\left(x\right) and p\left(x\right) are linear in x. If we define

<br /> q\left( p^{-1}\left(x\right) \right)=g\left(x\right)=ax+b<br />

and make the substitution

<br /> p\left(x\right) =\tilde{x}<br />

then

<br /> \int_{0}^{s}dx\frac{\sin q\left(x\right)}{1+p^{2}\left(x\right)}<br /> =\frac{1}{\left[p\left(1\right) -p\left(0\right)\right]}<br /> \int_{p\left(0\right)}^{p\left(s\right)}dx\frac{\sin g\left(x\right) }{1+x^{2}}=\frac{\left(2i\right)^{-1}}{\left[p\left(1\right)-p\left( 0\right) \right]}\sum_{\pm}\int_{p\left(0\right)}^{p\left( s\right) }dx\left(\frac{\pm e^{\pm ig\left(x\right)}}{1+x^{2}}\right)\,.<br />

If we examine just one of these integrals

<br /> \int_{p\left( 0\right) }^{p\left( s\right) }\frac{e^{ig\left( x\right)<br /> }}{1+x^{2}}=\int_{p\left( 0\right) }^{p\left( s\right) }dx\frac<br /> {e^{iax+ib}}{1+x^{2}}=ae^{ib}\int_{ap\left( 0\right) }^{ap\left( s\right)<br /> }dx\frac{e^{ix}}{a^{2}+x^{2}}\,,<br />

while the other is

<br /> \int_{p\left( 0\right) }^{p\left( s\right) }dx\frac{e^{-ig\left(<br /> x\right) }}{1+x^{2}}=\int_{p\left( 0\right) }^{p\left( s\right) }<br /> dx\frac{e^{-iax-ib}}{1+x^{2}}=-ae^{-ib}\int_{-ap\left( 0\right)<br /> }^{-ap\left( s\right) }dx\frac{e^{ix}}{a^{2}+x^{2}}\,.<br />

We know that

<br /> \int dx\frac{e^{ix}}{a^{2}+x^{2}}=\frac{i}{2a}\left[e^{-a}E_{1}\left(-a-ix\right) -e^{a}E_{1}\left( a-ix\right) \right] +C<br />

where E_n(x) is the exponential integral function (http://www.math.sfu.ca/~cbm/aands/page_230.htm relation 5.1.41)

Therefore

<br /> \int_{0}^{s}dx\frac{\sin q\left( x\right) }{1+p^{2}\left( x\right) }<br /> = [\ldots]<br />

which is just a big long expression in terms of E_1.

 

[pivoxa15]

Looks interesting although hard, I would like to see the exponential integral function representation.

Does the tidle x represent ax+b where a and b are any arbitary constants because that is what the form of p(x) should be in.

<br /> p\left(x\right) =\tilde{x}<br />

 

[jpr0]

Looks interesting although hard, I would like to see the exponential integral function representation.

Does the tidle x represent ax+b where a and b are any arbitary constants because that is what the form of p(x) should be in.

<br /> p\left(x\right) =\tilde{x}<br />

Yes, I just made a substitution, so the new integration variable is \tilde{x}, but because it's a dummy variable, i just renamed \tilde{x} as x because it gets tiresome typing tildes everywhere. Sorry for the confusion.

<br /> \tilde{x} = p(x) = cx+d<br />

where p(x) is just your linear function of x, and the term

<br /> \frac{1}{p(1)-p(0)} = \frac{1}{c}<br />

... this just comes from the change of the integration variable.

<br /> d\tilde{x} = \frac{dp(x)}{dx}dx = cdx = [p(1)-p(0)]dx<br />

I wrote it in this way because I didn't want to have a, b, c, d, and many more constants floating around everywhere. I just tried to write the limits of integration and prefactors in terms of you arbitrary linear functions of x. This is also why I defined the function g(x) ...
If

<br /> \tilde{x} = p(x)<br />

then

<br /> x = p^{-1}(\tilde{x})<br />

and

<br /> q(x) = q(p^{-1}(\tilde{x})) = g(\tilde{x})<br /> [/itex]<br /> <br /> which, after renaming \tilde{x} as x becomes g(x).<br /> And the limits of integration, when x=0, \tilde{x}=p(0), and when x=s, \tilde{x}=p(s). <br /> <br /> The integral as it is written above (my previous post) is <i>already</i> in terms of exponential integral functions. You just have to substitute your limits of integration... If you look in Abramovitz and Stegun then you will find (probably) all the information you need on these functions, like asymptotics and so on.

 

[Mindscrape]

Yeah, I guess I should have tried a little of it before suggesting parts. I figured it was a Calc2 problem where you could get an arctan and use some indentities to simplify and solve.

Did not expect the complex plane to enter into any of it! My bad.