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We know that,

1. $$ \int_0^1 t^{b-1} (1-t)^{c-b-1} (1-tz)^{-a} dt = \beta(b, c-b) ~\,_2F_1~(a, b, c, z) $$

2. $$ \int_0^1 t^{a-1} (1-t)^{b-1} e^{ct} dt = \beta(a, b) ~\,_1F_1~(a, b, c) $$ where $\beta(\cdot, \cdot)$, $\,_2F_1~(\cdot, \cdot, \cdot, \cdot)$, $~\,_1F_1~(\cdot, \cdot, \cdot)$ are Beta function, Gauss Hypergeometric function and Confluent Hypergeometric function respectively.

Then, how to find the following integrals?

3. $$ \int_0^{\theta} t^{b-1} (1-t)^{c-b-1} (1-tz)^{-a}~dt = ? $$ 4. $$ \int_0^{\theta} t^{a-1} (1-t)^{b-1} e^{ct}~dt = ? $$ where $0 < \theta < 1$.

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Fist of all, Welcome to the site !

The first antiderivative is given in terms of the Appell hypergeometric function of two variables $$I=\int t^{b-1} (1-t)^{c-b-1} (1-tz)^{-a}\,dt$$ $$I=t^b \left(\frac{F_1(b;b-c,a;b+1;t,t z)}{b}+\frac{t F_1(b+1;b-c+1,a;b+2;t,t z)}{b+1}\right)$$

The second one seems to be much more problematic (because of the exponential) and I should try a series expansion $$J=\int t^{a-1} (1-t)^{b-1} e^{ct}\,dt= \int t^{a-1} (1-t)^{b-1}\left( \sum_{n=0}^\infty \frac {c^n}{n!} t^n\right)\,dt$$ $$J=\sum_{n=0}^\infty \frac {c^n}{n!}\int t^{n+a-1} (1-t)^{b-1}\,dt=\sum_{n=0}^\infty \frac {c^n}{n!}\,B_t(a+n,b)$$

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  • $\begingroup$ Thank you @Claude Leibovici $\endgroup$ – saroja kumar Singh May 24 at 14:29

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