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- Thread starter oneomega
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- #1

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- #2

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Try the substitution t=e

- #3

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- #4

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I'd like to elaborate on D H's suggestion a little bit.

You should note that there is an explicit mention of [tex]e[/tex]. Also the possible answers include the square root of pi. Shouldn't that ring a bell?

At least, this depends on your level of education. Try searching for gaussian integral. This might clear some things up for you.

It does solve the integral check it. It took me about the size of a postcard and 1 minute to find the answer.

You should note that there is an explicit mention of [tex]e[/tex]. Also the possible answers include the square root of pi. Shouldn't that ring a bell?

At least, this depends on your level of education. Try searching for gaussian integral. This might clear some things up for you.

It does solve the integral check it. It took me about the size of a postcard and 1 minute to find the answer.

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- #5

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i'm not aware of gaussian integral. i'll definitely check it out. but, pls tell me, how did you come to conclusion that it can be solved by gaussian integral.

when DH suggested e^-z^2 . i thoght it just meant i needed to sub t= e^-z^2.

- #6

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Sure it does. There is a very simple relationship between the integral you asked about in the original post and the following integral:

[tex]\operatorname{erf}(x) \equiv \frac 2 {\sqrt{\pi}} \int_0^x e^{-t^2} dt[/tex]

In particular, there's a direct relationship between your integral and erf(∞).

Try as hard as you can and you will not be able to express either ##\int e^{-t^2}dt## or ##\int \frac{dt}{\sqrt{-\ln t}}## in terms of the elementary functions. Make all the u-substitutions you can think of. It won't work. Neither ##e^{-t^2}## nor ##\frac 1 {\sqrt{-\ln t}}## are integrable in terms of the elementary functions.

That does not mean these functions don't have an integral. It just means you can't express those integrals in terms of the elementary functions.

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