Why is this integral well defined?, how do I calculate it?$\int_0^\infty[(K+x)^v-x^v]^2dx$

Question

I assume that $K$ is a positive number, $v$ is a real number such that $v \in (-0.5,0.5)$.

I have the integral: $\int_{-\infty}^0[(K-s)^v-(-s)^v]^2ds$. By using substitution I have that this is equal to $\int_0^\infty[(K+x)^v-x^v]^2dx$.

Since $2v \in (-1,1)$, I don't get any problems around zero with the integral. What I am having problems with is when we go to infinity. Obviously there must be someway that $(K+x)^v$ cancels out with $x^v$.

One attempt is trying to calculte $\int_0^M[(K+x)^v-x^v]^2dx$, and let M go to infinity. I get that $\int_0^M[(K+x)^v-x^v]^2dx=\int_0^M(K+x)^{2v}dx-2\int_0^M(K+x)^vx^vdx+\int_0^Mx^{2v}dx$.

$\int_0^M(K+x)^{2v}dx=\frac{1}{2v+1}[(K+M)^{2v}-K^{2v}]$

$\int_0^Mx^{2v}dx=\frac{M^{2v+1}}{2v+1}$.

But I am having trouble with the part $-2\int_0^M(K+x)^vx^vdx$. Do you see how to integrate this?

Do you see a way of solving this integral, and seeing that it is well defined?

Answer 1

You do not need to do calculation. By using the Lagrange Mean Value Theorem $f(t)=t^v$ in $[x,x+K]$, one has $$ (K+x)^v-x^v=vK(x+\xi)^{v-1}, \xi\in(0,K).$$ Thus $$ ((K+x)^v-x^v)^2=v^2x^2(x+\xi)^{2v-2}=\frac{v^2K^2}{(x+\xi)^{2-2v}}\le \frac{v^2K^2}{x^{2-2v}}. $$ Noting $v\in(-0.5,0.5)$, one has $\int_1^{\infty}\frac{v^2K^2}{x^{2-2v}}dx$ converges and so does $\int_0^{\infty}[(K+x)^v-x^v]^2dx$.