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Does anyone know how to evaluate

$$I(a,b) = \int_0^1 B_t(a,b) dt$$?

Here $B_t(a,b)$ is the incomplete Beta function, defined as

$$B_t(a,b) = \int_0^t x^{a-1}(1-x)^{b-1} dx $$

Of course $I(a,b)$ equals

$$I(a,b) = \int_0^1 \int_0^t x^{a-1}(1-x)^{b-1} dx dt$$

but this does not help me very much. I tried to compute the derivatives $\displaystyle \frac {\partial I}{\partial a}$ but a logarithm appears and I can't get rid of it

Do you have any ideas?

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Intgration by parts, $$\eqalign{I(a,b)&=\left[(t -1)B_t(a,b)\right]_0^1+\int_0^1(1-t)t^{a-1}(1-t)^{b-1}dt\cr &=B(a,b+1)}$$

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  • $\begingroup$ thanks! It was actually easier than I thought :D $\endgroup$ – Ant Nov 11 '15 at 20:01

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