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**zetafunc.****Guest**

I've been trying to do

and

using DUIS, but I can't think of any kind of useful parametrisation that would work. Every time I do, I usually end up with something that *looks* like you can use integration by parts, but that doesn't work. I'm aware the indefinite integral form of these integrals can't be expressed in terms of elementary functions, so I'm hoping I might have more luck with the improper ones. Can anyone show me a useful parametrisation that would work here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Hi;

I have never seen it done using differentiation under the integral sign.

**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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**zetafunc.****Guest**

The last page here says you can do it.

tinyurl. com /bj99zrg

(remove spaces)

(I am assuming they meant cos(x[sup]2[/sup]) and not cos[sup]2[/sup]x because the latter is not a Fresnel integral and not very difficult.)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

**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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**zetafunc.****Guest**

I know you can do it without DUIS (e.g. gamma function method), but was curious if it was made really simple via DUIS (like integrating sinx/x).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Hi;

(I am assuming they meant cos(x^2) and not cos^2x because the latter is not a Fresnel integral and not very difficult.)

If you meant the examples at the end of that pdf then it meant

and not the integral you want.

**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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**zetafunc.****Guest**

But that is not a Fresnel integral. They said that the middle row contains two Fresnel integrals... so either they meant that it does not contain two Fresnel integrals, or they didn't parenthesise the powers properly...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Those are not Fresnel integrals so he made one of two mistakes. He did not put the square in the proper spot or he does not know what a Fresnel Integral is.

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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**zetafunc.****Guest**

Hmm. So, there might not be a way. I will still try to look for one however. It seems I never get to use this little tool.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

Hi;

I tried a couple of parametrizations but did not have any luck producing one that gets the known answer.

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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**zetafunc.****Guest**

What sorts did you try? I tried something of the form ln(bsinx[sup]2[/sup]) or similar, in the hope that I could cancel the trig term, but it did not work. Maybe trying to use a trig identity would fare better?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,032

I tried that one and also sin(x^b).

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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