# What are some good ways of visualizing the type of substitution of this integral.

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.

• Hm, I arrived at this simplification as well and saw that it was close to the beta. But I couldn't transform into a beta for some reason. Thanks. Does it scales to a beta? I still cannot see it. Jan 20 at 9:57
• @PabloCarneiroElias: rescale the variable. (Beware, there was a typo that I fixed.)
– user65203
Jan 20 at 10:03
• I see. Thanks. How about the substitution. Could you come up with that substitution? Thanks. Jan 20 at 10:05
• @PabloCarneiroElias: hem, I said it twice.
– user65203
Jan 20 at 10:06
• I see now. Thanks! Jan 20 at 10:08