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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

Hi;

The evaluation numerically of this integral came up in a discussion with anonimnystefy:

This one can resist normal methods quite well. When we say evaluate numerically we typically mean 10 - 12 digits with a programmable calculator and at least 100 when using a CAS.

To start, although any CAS known will gag on this problem they are able to do pieces of so it is to our advantage to split the integral like this.

The first integral is not too difficult and any CAS can do it to 100 digits.

That is about 105 digits or so. This is the hard part:

we notice that asymptotically as x grows large the denominator will act like x^2. The x^2 term will completely drown out the cos(x). That suggests the following chain of moves,

Looking at the RHS we see the same idea works again. As x grows large the denominator acts like x^4 completely drowning out the x^2 cos(x) term. So let's add a cos(x) / x^4 to it, we get

As x grows larger the denominator acts like x^6 completely drowning out the x^4 cos(x) term. So let's subtract a cos(x)^2 / x^6 to it, we get

Two things should be apparent,

1) we can continue the series on the left indefinitely and the RHS is getting smaller and smaller for each term on the left. The RHS will eventually be 0 as we get an infinite series on the LHS.

This becomes:

What have we accomplished? We have replaced a tough integral with a sum of infinite number of hopefully easier to evaluate numerically integrals. The beauty is that although we have an infinite number of integrals we will only need a finite number of them for the accuracy required.

Actually we will only need the first 26 or so terms.

Now I am not proposing this method of evaluating that integral because in practice those 26 integrals can also pose problems and further work is needed. But you wanted to see how the acceleration works.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

How many terms would you need for 1000 digits?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

Hi;

I estimate about 255 terms. As I said this method is too weak for that many digits. I just showed it because the transformation is one of the weapons used in numerical work.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Do sum accelerators work on it (like RRA and stuff...) ?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

I do not know the answer to that. I gave up on it when the 26 integrals were more difficult than they should have been.

Want to see another solution?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Of course!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

This method was showed to me.

Divide the integration region into lengths of

2 π and use cos(x) = cos(x + n*2π) having n as an integer. Then

now according to him we can interchange the order of the sum and the integral to get:

Mathematica can get the sum.

Now you just numerically integrate

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Have you been able to get 1000 digits out of that?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

I have not tried!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Why not?

And, it seems to me that the point is to get the integral into a form with finite integration limits...

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

For several reasons I do not like that solution.

1) I do not understand it!

2) Different versions of Mathematica were unable to do the sum.

3) I suspect it would take too long.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

What do you not understand?

What can Mathematica do with the integral

*Last edited by anonimnystefy (2013-02-03 02:40:57)*

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

The first line, beats me.

That will not work.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Well, if you do not understand it, there must be a part which you do not understand...

Why not?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

The trig substitution that produced the series, I do not get.

I tried a million substitutions to get the interval from 0 to infinity down to 0 to 1. Each one failed. Examine your integrand around 1.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

He just splits it up in intervals and uses cos(x+2n*pi)=cos(x). I do not see what the big deal is...

The integrand of my integral tends to 1 at 1.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

The cos(x) does not change, the x^2 does. Why?

There is a big fat indeterminate at x = 1 for your function. Numerical techniques will have problems with that.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Because cos(x+n*2pi)=cos(x).

Did you actually try entering the integral into Mathematica?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

Hi;

Because cos(x+n*2pi)=cos(x).

I am not getting that.

Yes, I entered it into the built in command and one I wrote. Both could not deal with it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Cosine has a period of 2pi, that's why how ever many times you add 2pi to the argument of cosine you will get the same thing.

Interestingly enough, WA gives 5 digits for that integral...

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

I know that one but I still do not undertsand what he did.

Probably that is all the digits could get. That integral is very difficult to compute.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Hi

What do you get when you substitute t=x+n*2pi into the integrals?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

I get a big mess:

I left out the intervals of integration because they remain the same.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

What do you get as the differential of t, i.e. as dt?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,469

I got dt = 2 π n dx

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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