# Improper Integrals ## Advanced problem #5 ### Andrew Freeman 11-13-20 live: https://hackmd.io/@AEjIFj77Rxyj8Izesy-cPw/rkM_-cntw#/ --- $$B(a,b)=∫^1_0t^{(a−1)}(1−t)^{(b−1)}dt$$ In this section you are asked to show that this integral converges as long as a>0 and b>0. The problem lies when a<1 or b<1 as in those cases one of the powers is negative and therefore the integral is improper in either the 0 endpoint or the 1 endpoint. Show that the integral does in fact converge for such values. --- The first step is to split the integral in two. $$B(a,b)=∫^1_0t^{(a−1)}(1−t)^{(b−1)}dt=$$ $$\int^\frac{1}{2}_0t^{(a−1)}(1−t)^{(b−1)}dt-\int^1_\frac{1}{2}t^{(a−1)}(1−t)^{(b−1)}dt$$ --- These integrals are equal to each other, so if it is found that one of them converges it can be said that they both converge. --- Using theorem two from this section it can be found that one of them converges. Below is the theorem for when $p<1$. $$\int^a_0\frac{dx}{x^p}=\frac{a^{1-p}}{1-p}$$ --- In our case $p=0$ and $p=-1$ in order to show that it converges when a, b both equal 0 or 1. $$\int^\frac{1}{2}_0t^{(a−1)}(1−t)^{(b−1)}dt=\frac{a^{1-0}}{1-0}$$ $$\int^\frac{1}{2}_0t^{(a−1)}(1−t)^{(b−1)}dt=\frac{a^{1--1}}{1--1}$$ --- In both of those situations $p<1$, so it is converging, and since both halfs of the integral are the same, they both converge. Since both of the halfs are converging, then the larger whole integral must also be converging. ---