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Someone has an idea to calculate the following integral

$$I_{\alpha} = \int_{0}^{+\infty} t^{-\alpha} (1-a)^{t} dt; \quad 0<a,\alpha<1.$$ Thank you in advance

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this is $\int_0^\infty e^{-bt}t^{(1-\alpha)-1}dt$ where $b=-\log (1-a)>0$. Substitute $bt=u$ you will get Gamma function.

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