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I found the following formula in "INTEGRALS AND SERIES, vol.3" by Prudnikov, Brychkov and Marichev (page 188, eq.5).

$$\int_0^{\infty} \frac{x^{\alpha-1}}{\sqrt{(a+x)^2+z^2}}K(\frac{2\sqrt{ax}}{\sqrt{(a+x)^2+z^2}})dx = \frac{\sqrt{\pi}}{4}\Gamma(\frac{\alpha}{2})\Gamma(\frac{1-\alpha}{2})(a^2+z^2)^{(\alpha-1)/2}P_{-\alpha}(\frac{z}{\sqrt{a^2+z^2}})$$

$K$: complete elliptic integral of the first kind.

$\Gamma$: Gamma function

$P$: Legendre function of the first kind.

$\alpha$: complex value

The book says this is valid for $a>0, Re\{z\}>0, 0 < Re\{\alpha\}<1$. Could anybody tell me how this can be derived ?

I think this formula would be useful for evaluating Mellin transform of a function that contains the complete elliptic integrals.

Thank you in advance.

https://www.researchgate.net/publication/268650078_Integrals_and_Series_Volume_3_More_Special_Functions

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    $\begingroup$ 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.) $\endgroup$
    – Kazuki OKAMURA
    Commented Feb 1, 2023 at 13:41
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    $\begingroup$ 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. $\endgroup$
    – r-nishi
    Commented Feb 1, 2023 at 14:10
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    $\begingroup$ Answer was given at math.stackexchange.com/a/4632164/1145590 Thank you for cooperation. $\endgroup$
    – r-nishi
    Commented Feb 6, 2023 at 7:00

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