It's a homework task and I can't get past the last step.

Task is to prove that $$ B(x,y)=\int\limits_0^1 \frac{\tau^{x-1}+\tau^{y-1}}{(1+\tau)^{x+y}} \mathrm{d}\tau $$

By substituting $t=\frac{1}{\tau+1}$ into beta integral $$ B(x,y)=\int\limits_0^1 t^{x-1}(1-t)^{y-1}\mathrm{d}t $$

and repeating it with substition $t=\frac{\tau}{\tau+1}$ and counting both results together it can be shown that $$ 2B(x,y)=\int\limits_0^{+\infty} \frac{\tau^{x-1}+\tau^{y-1}}{(1+\tau)^{x+y}} \mathrm{d}\tau $$

Can this result be transformed into the needed one? Or should the initial steps be different?


Hint: $$\int_1^{+\infty}\frac{\tau^{x-1}}{(1+\tau)^{x+y}}\mathrm d\tau\ \stackrel{\sigma=1/\tau}{=}\ \int_0^1\frac{\sigma^{y-1}}{(1+\sigma)^{x+y}}\mathrm d\sigma$$ Thus the initial steps you show are all right.

  • $\begingroup$ Thanks, didn't see that :) $\endgroup$ – Džuris Jun 3 '12 at 14:42

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