Talk:List of integrals of exponential functions

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1 ?
2 Nicer integral
3 Inaccurate Integral
4 Needs proofs
5 one more please
6 beware

[edit] ?

Removed this, since it is incorrect:

$\int_{-\infty}^{\infty} e^{-{x^2}/{a^2}} \cos bx\,dx=a \sqrt{\pi}(\sin{a^2 b^2 \over 4}+\cos{a^2 b^2\over 4})$

Try numerically with $a = b = 1$ . Should probably be ${a\sqrt{\pi}} \exp(-({ab \over {2}})^2)$ for $a > 0$ . TB

This integral doesn't make much sense; what's n?

$\int e^{x^2}\,dx = e^{x^2}\left (\sum_{r=1}^n\frac{1}{2^n x^{2n-1}} \right)+ \frac {2n-1}{2^n}\int \frac{e^{x^2}\;dx}{x^{2n}}$

Ed Sanville 00:23, 30 May 2005 (UTC)

I guess it should hold with any n. I think this is obtained by integrating by parts n times. So, I don't know the details, but I have a feeling this formula is correct as it stands. Oleg Alexandrov 22:04, 1 Jun 2005 (UTC)

Yeah, it does appear to be the n-th iteration of integration by parts. I think it's mathematical common sense that if nothing is said about n then you get to pick your favourite one, but for the people with actual common sense, we should probably include some comment which tells them that. I know this problem has come up regarding this specific integral among my friends. I just don't think I'm the right person to put that blurb in, for fear of confusing them further. Dusik 16:58, 22 September 2005 (UTC)

The formula is in fact obtained by repeated application of partial integration but is not correct as it stands (wrong coefficients). The (presumably) correct formula can or could be derived from the asymptotic expansion of the error function and is (to be checked once more)

$\int e^{x^2}\,dx = e^{x^2}\left( \sum_{j=0}^{n-1}\frac{2j\,!}{j!} \,\frac{1}{2^{2j+1}x^{2j+1}} \right )+\frac{(2n-1)\,!}{(n-1)!2^{2n-1}}\int \frac{e^{x^2}}{x^{2n}}\;dx \quad \mbox{ for } n > 0$

I thought that the above integral was not convergent. Is the solution exact?

In the case of the errorfunction (integral : $\int_{0}^{t} e^{-x^2}\,dx$ ) the expansion has the same coefficients but now with alternating sign. The main application of this integral identity is to give an approximation formula (for large t) for the error function values or the complementary (erfc-) function. Here one has to choose a suitable n (cut-off parameter) depending on t because the full series diverges for every x, it is only an asymptotic series.

--212.18.24.11 09:14, 23 September 2005 (UTC)

[edit] Nicer integral

ive found a nicer looking solving of the integral

$\int x^n e^{cx}\; dx = \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} dx$

wich to me looks nasty. i have a easier method. should i add it?

[edit] Inaccurate Integral

$\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a^3}$

is incorrect. That's the answer if the limits were

$\int_{0}^{\infty}$

I've changed it to the correct:

$\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\sqrt{\pi \over a^3}$

Melink14 18:43, 27 February 2007 (UTC)

I lied. It was correct as before as far as I can tell. If you can verify please do. Melink14 19:21, 27 February 2007 (UTC)

[edit] Needs proofs

The page needs proofs. That is, for every identity a link to a proof should be included, perhaps by linking to a page which contains proofs directly, like:

<complicated integral> = <nice value> (proof)

Or:

<complicated integral> = <nice value> (proof)

Or perhaps using <ref>:

<complicated integral> = <nice value> [1]

To me most of them look doable, but I assume that to a layman this looks just like your average unreferenced article. Shinobu 23:27, 1 June 2007 (UTC)

[edit] one more please

can someone please verify this, and add to the table?

$\int {x \over \sigma\sqrt{2\pi} }\,e^{-{(x-\mu )^2 / 2\sigma^2}}\; dx= \frac{1}{2} \mbox{erf}\,\frac{x-\mu}{\sigma \sqrt{2}}-\frac{\sigma}{\sqrt{2\pi}}\,e^{-{(x-\mu )^2 / 2\sigma^2}}$