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mooeypoo

Integrals + completing the square

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Hey guys,

 

I'm working on a fourier transform and got stuck in a stage where I think I'm supposed to complete the square .. but.. I don't manage to do it.

 

Here's what I have:

 

[math]\int \exp{(\frac{-\alpha t^2}{\triangle t^2})} \exp{(-i\omega t)}=\int \exp{-(\frac{\alpha t^2}{\triangle t^2}+i\omega t)}[/math]

 

And so from this point I want to complete the square so I can continue to solve. Here's how I started:

 

[math]\frac{\alpha}{\triangle t^2}\left( t^2 + \frac{i\omega \triangle t^2}{\alpha}t \right) = \frac{\alpha}{\triangle t^2}\left( t^2 + \frac{i\omega \triangle t^2}{\alpha}t + (\frac{i\omega \triangle t}{2\alpha})^2 - (\frac{i\omega \triangle t}{2\alpha})^2 \right) [/math]

[math]= \frac{\alpha}{\triangle t^2} \left[ \left(t + \frac{i\omega \triangle t^2 }{\alpha} \right)^2- (\frac{i\omega \triangle t}{2\alpha})^2 \right] [/math]

 

So that's what I have now in the exponent:

 

[math]\int \exp{\frac{\alpha}{\triangle t^2} \left[ \left(t + \frac{i\omega \triangle t^2 }{\alpha} \right)^2- (\frac{i\omega \triangle t}{2\alpha})^2 \right]}=[/math]

 

But.. now what? Our professor told us that the way to go from the first stage is to complete the square and use an integral table to continue, but I don't see how that helps me...

 

 

What did I do wrong? Any ideas?

 

Thanks!!

 

~moo

 

 

p.s-- i accidentally wrote t instead of t^2 in the first instance, which made the first equation wrong.. I fixed it now. The equation is right, the missing square was a typo.

Edited by mooeypoo

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What did I do wrong? Any ideas?

Firstly, you did not specify the variable of integration, which I assume is [math]t[/math].

 

Secondly, you used the fact that [math]\exp(a)\exp(b) = \exp(a+b)[/math] for the very first step. That's fine. What's wrong is that was use that this relation works both ways. Use this relation again, but in reverse.

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Firstly, you did not specify the variable of integration, which I assume is [math]t[/math].

 

Secondly, you used the fact that [math]\exp(a)\exp(b) = \exp(a+b)[/math] for the very first step. That's fine. What's wrong is that was use that this relation works both ways. Use this relation again, but in reverse.

Oh, shoot, you're right, sorry, yes all integrals should end with "dt".

 

Sorry 'bout that.

 

And... hm. Okay... trying now.


Merged post follows:

Consecutive posts merged

Okay, so I have one more step:

[math]\int \exp{\frac{\alpha}{\triangle t^2} \left[ \left(t + \frac{i\omega \triangle t^2 }{\alpha} \right)^2- (\frac{i\omega \triangle t}{2\alpha})^2 \right]} dt = \int \exp{\left[ \frac{\alpha}{\triangle t^2} \left(t + \frac{i\omega \triangle t^2 }{\alpha} \right)^2 - \frac{i^2\omega^2}{4\alpha} \right]} dt [/math]

 

And then "reversing" the process:

[math] \int \frac{ \exp{ \frac{\alpha}{\triangle t^2} ( t + \frac{i\omega \triangle t^2 }{\alpha} )^2 }}

{\exp{\frac{i^2\omega^2}{4\alpha}} } dt[/math]

 

But.. how does that help me? :-( it seems to make it more complicated to solve, not easier...

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The denominator is a constant. It is just a number. You can take it outside the integral.

 

BTW, what is [math]i^2[/math] ?

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BTW, what is [math]i^2[/math] ?

 

[hide]It's -1[/hide]

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omg i feel silly, I didn't notice the fact it's a constant!!! bah.

 

And yes, I know i^2=-1, I forgot to change it.

 

Here's what I have, then:

 

[math]

\int \frac{ \exp{ \frac{-\alpha}{\triangle t^2} ( t + \frac{i\omega \triangle t^2 }{\alpha} )^2 }} {\exp{\frac{i^2\omega^2}{4\alpha}} } dt = \exp{\frac{\omega^2}{4\alpha}} \int \exp{\frac{-\alpha}{\triangle t^2} \left( t+\frac{i\omega\triangle t}{2\alpha} \right) ^2 } dt

[/math]

 

BTW, <sigh> if you go back to the initial equation, alpha was negative, and I seem to have neglected that at some point. It's back on the equation above, though.

 

In any case, I looked at the table of integrals and I can't really find a way to solve this in the indefinite integral.

 

I was thinking of continuing with a u substitution:

 

[math]u=t+\frac{i\omega\triangle t}{2\alpha}[/math]

[math]du=dt[/math]

 

And therefore:

 

[math]

\exp{\left( \frac{\omega^2}{4\alpha} \right) } \int \exp{\left( \frac{-\alpha}{\triangle t^2} u^2 \right) }du=

[/math]

 

But now what?

 

I went over the table of exponential integrals (the professor actually suggests we do this in the hw sheet) and the only solution I can see is if my integral is definite, which it shouldn't be, because it's a fourier transform.

 

There's also this: 41e064bfa4da741bc25cfab7793040cf.png

But what is this "error function"? Is this really it? It's the first time I see anything like that, so I'm not sure.

 

Gah!?

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In any case, I looked at the table of integrals and I can't really find a way to solve this in the indefinite integral.

Whoa, there!

 

You made another very basic mistake in the original post. I assume you are trying to find the fourier transform of [math]\exp(-at^2/\Delta t^2)[/math]. The correct way to do this is via the definite integral

 

[math]\int_{-\infty}^{\infty}

\exp(-at^2/\Delta t^2)\,\exp(-i \omega t)\,dt[/math]

 

Note very well: This is a definite integral, not an indefinite integral.

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Whoa, there!

 

You made another very basic mistake in the original post. I assume you are trying to find the fourier transform of [math]\exp(-at^2/\Delta t^2)[/math]. The correct way to do this is via the definite integral

 

[math]\int_{-\infty}^{\infty}

\exp(-at^2/\Delta t^2)\,\exp(-i \omega t)\,dt[/math]

 

Note very well: This is a definite integral, not an indefinite integral.

... that's a definite integral???... uh.. really?

 

So.. uhm

7ea2d787aa8db58a979ce9a5ae35831d.png

 

 

Okay, so:

[math]

=\exp{\frac{\omega^2}{4\alpha}}\left( \sqrt{\frac{\pi \triangle t^2}{\alpha}} \right)

[/math]

 

And since [math]\omega = 2\pi \nu[/math]

 

So:

[math]

=\exp{\frac{\pi^2\nu^2}{\alpha}}\left( \sqrt{\frac{\pi \triangle t^2}{\alpha}} \right)

[/math]

 

And my initial alpha represented the constant 4ln2, so my final answer:

 

[math]

I(\nu)=\sqrt{\frac{\pi \triangle t^2}{4ln2}} \exp{\left( \frac{\pi^2\nu^2}{4ln2} \right)}

[/math]

Which is ugly, but should work.

 

Thanks :)


Merged post follows:

Consecutive posts merged

And now, just to verify, the next problem section asks to find

[math]\triangle \nu \triangle t = 0.44[/math]

Which is the gaussian value.

 

So to find the [math]\triangle t[/math] I am looking for I(t)=1/2 for the FWHM.

 

[math]I(t)=1/2=\exp{(\frac{-4ln2t^2}{\triangle t^2})}[/math]

[math]0-ln2=-\frac{4ln2t^2}{\triangle t^2}[/math]

[math]t^2=\frac{\triangle t^2}{4}[/math]

[math]t=\frac{\triangle t}{2}[/math]

[math]\triangle t=\frac{t}{2}[/math]

 

Which is the obvious solution..

 

And for [math]\triangle \nu[/math] I am looking for I(v)=1/2:

 

[math]I(\nu)=1/2=\sqrt{\frac{\pi \triangle t^2}{4ln2}}\exp{(\frac{\pi^2\nu^2}{4ln2})}[/math]

[math]\frac{\sqrt{ln2}}{\sqrt{\pi}\triangle t}=\exp{(\frac{\pi^2\nu^2}{4ln2})}[/math]

[math]ln \left( \frac{\sqrt{ln2}}{\sqrt{\pi}\triangle t} \right)=\frac{\pi^2\nu^2}{4ln2}[/math]

[math]\frac{4ln2}{\pi^2} ln \left( \frac{\sqrt{ln2}}{\sqrt{\pi}\triangle t} \right)=\nu^2[/math]

 

[math]\nu = \frac{2\sqrt{ln2}}{\pi}\sqrt{ln\left( \frac{\sqrt{ln2}}{\sqrt{\pi} \triangle t} \right)}[/math]

 

Logically, delta-nu would be double the above.

 

And so, technically, I'm supposed to show that

[math]

\triangle \nu \triangle t = 0.44

[/math]

 

Which is a hugeomongous headache, but should be the right method... I think.

Edited by mooeypoo

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... that's a definite integral???... uh.. really?

Really. Look in your text.

 

So.. uhm

[math]\int_{-\infty}^{\infty}e^{-ax^2} dx = \sqrt{\frac{\pi} a}[/math]

That's correct, but you have

 

[math]

\int_{-\infty}^{\infty} \exp\left({\frac{-\alpha}{\triangle t^2} \left( t+\frac{i\omega\triangle t}{2\alpha} \right) ^2 }\right) dt

[/math]

 

To make the above less messy, call it

 

[math]\int _{-\infty}^{\infty} \exp\left(-(at+ib)^2\right)\,dt[/math]

 

It is easy to go from the Gaussian integral

 

[math]\int _{-\infty}^{\infty} \exp\left(-t^2\right)\,dt = \sqrt{\pi}[/math]

 

to

 

[math]\int _{-\infty}^{\infty} \exp\left(-at^2\right)\,dt

= \sqrt{\frac{\pi} a}[/math]

 

and even to

 

[math]\int _{-\infty}^{\infty} \exp\left(-(at+b)^2\right)\,dt

= \sqrt{\frac{\pi} a}[/math]

 

All you need is a change of variables.

 

Showing that [math]\int _{-\infty}^{\infty} \exp\left(-(at+ib)^2\right)\,dt

= \sqrt{\frac{\pi} a}[/math]

 

is going to take a bit more work. The change of variables technique is not valid here. If this is for a physics class you can probably get away with it. If it is for a math class, you can't. Showing that the same result pertains with a complex offset requires the use of Cauchy's integral formula.

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My u-substitution won't work here, D H?

 

[math]u=t+\frac{i\omega\triangle t}{2\alpha}[/math]

[math]du=dt[/math]

 

And therefore:

 

[math]

\exp{\left( \frac{\omega^2}{4\alpha} \right) } \int \exp{\left( \frac{-\alpha}{\triangle t^2} u^2 \right) }du

=\exp{\left( \frac{\omega^2}{4\alpha} \right) }\left( \sqrt{\frac{\pi \triangle t^2}{\alpha}} \right)

[/math]

 

?

 

 

 

p.s: This is for a physics class.. specifically Optics.

Edited by mooeypoo

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Nope. It is not valid. When you are using a u-substitution in a definite integral you have to apply the substitution to the integration limits as well as to the integrand. The u-substitution works for a real offset because, informally, infinity plus some finite real offset is still just infinity. Infinity plus some finite imaginary offset is not just infinity.

 

There are many definite integrals where pretending the complex offset isn't going to affect the answer will get you in loads of trouble. In this case ignoring that the offset is complex turns out to yield the correct answer. The reason is because (a) exp(-z^2) tends to zero as Re(z) tends to plus or minus infinity, and (b) exp(-z^2) is analytic on the complex plane.

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Nope. It is not valid. When you are using a u-substitution in a definite integral you have to apply the substitution to the integration limits as well as to the integrand.

Ah! I thought of that, though, and infinity, no matter what I multiply, add or subtract to it, is still infinity. Same goes with negative inifinity.. so I figured the limits are okay to remain the same.

 

The u-substitution works for a real offset because, informally, infinity plus some finite real offset is still just infinity. Infinity plus some finite imaginary offset is not just infinity.

..wait, what?

 

Infinity + a constant is not infinity?........ really?...... what... is it?

 

There are many definite integrals where pretending the complex offset isn't going to affect the answer will get you in loads of trouble. In this case ignoring that the offset is complex turns out to yield the correct answer. The reason is because (a) exp(-z^2) tends to zero as Re(z) tends to plus or minus infinity, and (b) exp(-z^2) is analytic on the complex plane.

I'm not sure I understood the last bit (b), but I also never studied this. Where can I read a bit about these concepts?

 

You know, I already complained about it to my physics advisor (and to the Department Chair) but this is ridiculous. As a physics undergrad, I took math courses along with Engineers in courses that seemed to be tailored for Engineers. It might sound pretentious, but this is a serious drawback. Engineering undergrads' attitude towards math is to memorize the formula and use it when needed, as opposed to actually going over the (yeah, sometimes complex, boohoo) method of deriving the formula so that when I encounter something that has no formula, I can actually deal with it.

 

I don't think Engineering students know about the concept that sometimes things are.. like.. new.. and don't have formula.. and you need to actually use your logic and math skills to handle them.. gee.

 

Now that I go over actual Physics, in the course and in my research project I have to study most of the math myself. It's just annoying as hell, I feel so math-ignorant for a physics girl.

 

Anyhoo, I had to rant.

 

~moo

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Infinity + a constant is not infinity?........ really?...... what... is it?

I did say "informally". Formally, improper integrals such as

 

[math]\int_{-\infty}^{\infty}\exp(-x^2)\,dx[/math]

 

are better treated as a limit of a proper integral

 

[math]\lim_{L\to\infty}\int_{-L}^{L}\exp(-x^2)\,dx[/math]

 

Now add a complex offset to the integration interval:

 

[math]\lim_{L\to\infty}\int_{-L+ib}^{L+ib}\exp(-z^2)\,dz[/math]

 

Informally, this becomes

 

[math]\int_{-\infty+ib}^{\infty+ib}\exp(-z^2)\,dz[/math]

 

Not that that helps much because ...

I'm not sure I understood the last bit (b), but I also never studied this. Where can I read a bit about these concepts?

You're doing Fourier analysis and you haven't taken complex analysis?

 

Shudder.

 

Where can you learn it? As far as a cogent presentation goes, a book. Online stuff is too haphazard. I have Carrier, Krook, and Pearson, functions of a complex variable: theory and technique and Rudin, Real and Complex Analysis. Carrier is aimed at undergrads. Rudin is a lower level graduate text. Both are old, although Rudin is still apparently the gold standard.

 

[Rant mode elided] Anyhoo, I had to rant.

I hear ya. I'm a physicist by training but an aerospace engineer by profession. Sometimes the engineers I work with day in and day out can drive me nuts -- and these are aerospace engineers, one of the hardest of engineering disciplines.


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This page shows how to compute a Gaussian integral with a complex offset: http://ccrma.stanford.edu/~jos/sasp/Gaussian_Integral_Complex_Offset.html

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