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Recently I need to evaluate an integral looks like

$I=\int _0^1 dx \int_0^1 dy \frac{1}{(1-x)^{\frac{1}{2}}\left(y(1-xy)t+1\right)}$

where $1>t>0$.

I don't know how to do it, and I'm told to try Mellin-Barnes representation, which lead me to

$I=\frac{1}{(2\pi i)}\int _0^1 dx \int_0^1 dy\int _{-i\infty}^{i\infty}dz_1(1-x)^{-\frac{1}{2}} \Gamma (1+z_1)\Gamma (-z_1)\frac{(-xy^2t)^{z_1}}{(1+yt)^{1+z_1}}\\ =\frac{1}{(2\pi i)^2}\int _0^1 dx \int_0^1 dy\int _{-i\infty}^{i\infty}dz_1\int _{-i\infty}^{i\infty}dz_2 (1-x)^{-\frac{1}{2}}\\ \times \Gamma (1+z_1+z_2)\Gamma (-z_1)\Gamma (-z_2)(-xy^2t)^{z_1}(yt)^{z_2}\\ =\frac{\Gamma (\frac{1}{2})}{(2\pi i)^2}\int _0^1 dx \int_0^1 dy\int _{-i\infty}^{i\infty}dz_1\int _{-i\infty}^{i\infty}dz_2 (-1)^{z_1}t^{z_1+z_2}\\ \times \Gamma (1+z_1+z_2)\Gamma (-z_1)\Gamma (-z_2)\frac{\Gamma (2z_1+z_2+1)}{\Gamma (2z_1+z_2+2)}\frac{\Gamma (z_1+1)}{\Gamma (z_1+\frac{3}{2})}$

I think I can close the contour to the right and thus have (I'm not sure) $I=\Gamma (\frac{1}{2})\sum _{n_1=0}^{\infty}\sum _{n_2=0}^{\infty}\frac{(-1)^{n_1}(-t)^{n_1+n_2}}{n_1!n_2!}\Gamma (1+n_1+n_2)\frac{\Gamma (2n_1+n_2+1)}{\Gamma (2n_1+n_2+2)}\frac{\Gamma (n_1+1)}{\Gamma (n_1+\frac{3}{2})}$

However, it seems that, because of $\frac{\Gamma (1+n_1+n_2)}{n_1!n_2!}$, the sum dose not converge. I also tried to throw it to Mathematica, but failed to obtain a result.

I hope my understanding of the Mellin-Barnes representation is correct. I need help.

By the way, if there is another way to evaluate the original integral, it is also interesting!

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After a long time this is not answered. However I'm able to deal with this using another method.

First substitude $x\to 1-x$, the integral is $$\int _0^1dx\int _0^1dy\frac{1}{\sqrt{x}(1+ty-ty^2+txy^2)}$$ Then $x\to x^2$ it is $$2\int _0^1dx\int _0^1dy\frac{1}{(1+ty-ty^2+tx^2y^2)}$$ Then $x\to \frac{x}{y}$ it is $$2\int _0^ydx\int _0^1dy\frac{1}{y(1+ty-ty^2+tx^2)}$$ Then $x\to x+y$ it is $$2\int _{-y}^0dx\int _0^1dy\frac{1}{y(1+ty+tx^2+2xyt)}$$ Then $x\to -x$ $$-2\int _0^ydx\int _0^1dy\frac{1}{y(1+ty+tx^2-2xyt)}$$ Now, change the order of the integral, it is $$-2\int _0^1dx\int _x^1dy\frac{1}{y(1+ty+tx^2-2xyt)}$$ Now I am able to throw it to Mathematica, and the result can be obtained.

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