I came across the given integral while solving a problem:
$$\int_u^{1} v^{\beta-1} (1-v)^{\gamma-1}\left(\frac{v-u}{v}\right)^{\beta-1}dv$$
I couldn't find a way of simplifying it at all using substitution. After giving up I checked some proposed hints and they showed up this substitution:
$$ y = \frac{v-u}{1-u}, dy= \frac{dv}{1-u} $$
Then the integral simplifies to a beta:
$$\int_0^{1} y^{\beta-1} (1-y)^{\gamma-1}dy$$
which is known.
I known the beta integral very well and I am aware that it was probably the best solution to try to get to a known integral from the original one. However, I still don't know a systematic way of coming up with this substitution. Seems out of the blue.
Can someone help developing techniques to be able to quickly visualize this kind of substitution?
Edit: I understand that substitution technique comes from the "reversed chain rule" in integration. But in this case I couldn't see the pattern. The real question is: how to develop some technique to be able to come up to this substitution? Or also, in this case, what are the main patterns in the integral that would suggest this substitution?
Thanks a lot.
Mh, your example is somewhat rigged, because
$$ v^{\beta-1} (1-v)^{\gamma-1}\left(\frac{v-u}{v}\right)^{\beta-1}$$ immediately simplifies as
$$ (1-v)^{\gamma-1}\left(v-u\right)^{\beta-1}.$$
Then $t:=1-v$ brings you very close to a Beta:
$$ t^{\gamma-1}\left((1-u)-t\right)^{\beta-1}.$$ Just rescale.