# Calculate $I(\alpha, x,y)=\int\limits_0^1 {{v^{\alpha - 1}}{{(1 - vx)}^{\alpha - 1}}{e^{vy}}dv,\,\,\,0 < \alpha ,x,y < 1}.$

I want to calculate this integral with singularity:

$$I(\alpha, x,y)=\int\limits_0^1 {{v^{\alpha - 1}}{{(1 - vx)}^{\alpha - 1}}{e^{vy}}dv,\,\,\,0 < \alpha ,x,y < 1}.$$

I hope to obtain a closed formula via special functions, maybe some hypergeometric functions. A related differential equation for solving $I(\alpha, x,y)$ as a function of $x$,$y$ is also interesting.

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Expand in series with respect to $y$ and integrate term-wise: $$I(\alpha, x,y) = \sum_{n=0}^\infty \frac{y^n}{n!} \int_0^1 v^{n+\alpha-1} (1-x v)^{\alpha-1} \mathrm{d} v = \sum_{n=0}^\infty \frac{y^n}{n!} \frac{1}{n+\alpha} {}_2 F_1\left(n+\alpha, 1-\alpha; n+\alpha+1 ;x\right)$$ From here you see that this is confluent bivariate hypergeometric function, Horn function $\Phi_1$: $$I(\alpha, x, y) = \frac{1}{\alpha} \Phi_1\left(\alpha, 1-\alpha, \alpha+1, x, y\right)$$