Timeline for Derivation of an integral containing the complete elliptic integral of the first kind
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2023 at 7:00 | comment | added | r-nishi | Answer was given at math.stackexchange.com/a/4632164/1145590 Thank you for cooperation. | |
| Feb 1, 2023 at 14:12 | history | edited | r-nishi | CC BY-SA 4.0 |
added 10 characters in body
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| Feb 1, 2023 at 14:10 | comment | added | r-nishi | Thank you for the comment. Your comment is right. There was a typo. $\Gamma(\frac{a}{2}) \Gamma(\frac{1-a}{2})$ --> $\Gamma(\frac{\alpha}{2})\Gamma (\frac{1-\alpha}{2})$. I changed the original question, accordingly. | |
| Feb 1, 2023 at 13:41 | comment | added | Kazuki OKAMURA | Let $w = z/a$. Then, $\text{Re}(w) > 0$. Consider a change of variable $y = x/a$ in the integral. Then, the formula is equivalent with $\int_0^{\infty} \frac{y^{\alpha-1}}{\sqrt{(1+y)^2 + w^2}} K(\frac{2\sqrt{y}}{\sqrt{(1+y)^2 + w^2}}) dy = \frac{\sqrt{\pi}}{4} \Gamma(\frac{a}{2}) \Gamma(\frac{1-a}{2}) (1+w^2)^{(\alpha - 1)/2} P_{-\alpha}\left(\frac{w}{\sqrt{1+w^2}}\right)$. However, $\Gamma(\frac{a}{2}) \Gamma(\frac{1-a}{2})$ is not a constant function for $a$, so there is something wrong in the formula. (For the square root of complex numbers, we assume a usual branch cut.) | |
| Jan 31, 2023 at 10:12 | history | edited | r-nishi | CC BY-SA 4.0 |
edited title
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| S Jan 31, 2023 at 9:42 | review | First questions | |||
| Jan 31, 2023 at 10:51 | |||||
| S Jan 31, 2023 at 9:42 | history | asked | r-nishi | CC BY-SA 4.0 |