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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi,

I was trying a problem from AMM. But to proceed, I need to know whether this integral can be done by hand:

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi gAr;

I do not think so. Not the indefinite one.

You might be able to do it because it is a definite integral.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

Okay.

Even if I could prove it, I end up with an infinite product, even more difficult!

The integral I'm trying is in this page: http://www.mat.uniroma2.it/~tauraso/AMM/amm.html

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi;

What problem number is it? What I am doing is looking through a very large table of integrals. If someone has ever done that or a similar one before it might show up there.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

It's "Problem 11564 - 04 - By A. Stadler (Switzerland)."

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Nothing yet gAr.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Okay, no problem. I tried expanding, substitution etc., but anyway not an easy one.

That guy is very good, solving most of the problems from the monthly!

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi gAr;

That problem I believe was not solved by anyone.

Anyway it does not look like that integral has an elementary form. Here is why.

The above integral does not have a closed form according to Table of Integrals, Series and Products 7th edition p 390. They give a series for the answer, so no closed form. It is easy to see that your integral can be split into 2 integrals that are of that same form.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

Okay, thanks for the info.

Can you evaluate my first post from the series given in the book?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi gAr;

I believe so. There a several simpler forms. They do not involve a series but all the answers contain the Ei(x) form. This is called the exponential integral. To save time it was tabulated and named. But it is an integral, so it is not an elementary form.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Okay.

I tried Riemann sum and it became even more difficult!

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

What I do not like is that when I split the numerator, suddenly the first integral no longer converges. That x in the denominator is now a problem.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Yes, that's a big problem.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi;

Here is what they have for integral 3.354 4

I need to put beta = 0 but there are conditions.

Yikes! I just saw it now, it zeroes the argument of the exponential integral! Beta can not be 0.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Is there any of the form:

?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

I am still looking, if it does, then you could use it. A table is a valid way to do an integration.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Computationally, it's easy to verify. But I can't derive it.

I can't search for such equations since search engines ignore the mathematical symbols. Bad!

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Bingo!

3.434 2

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

That's cool!

Does it mention anything more about the integral?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi;

No it does not. It is just a table of tons of them. Most of the ones known.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hmmm, it would be nice if I could get a way to derive it.

Anyway, now I'm left with the product

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi gAr;

You can look it up again!?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Yeah, but is it a good way to solve a problem?!

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Depends. If you are working on something larger than you do not want to get tied up on a tiny detail. Here, it is not helpful to use a table or a package. But what else is there?

I might be able to change that to a sum but it is more difficult.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Evaluating that product would solve the integral.

So the only part which I did here is series expansion.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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