TITLE: How to evaluate the integral $\int_1^{\infty}[u^\alpha-(u-1)^{\alpha}]^2du$?
QUESTION [7 upvotes]: The integral
\begin{equation}
I=\int_1^{\infty}[u^\alpha-(u-1)^{\alpha}]^2du,\quad -1/2<\alpha<1/2
\end{equation}
comes from the stochastic process fractional Brownian motion.
Can someone show how to evaluate it?
Thank you!
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Reply to Lucian:
Do you mean the following derivation?
By letting $t=1/u$, the integral becomes
\begin{equation}
\begin{split}
I&=\int_0^1\left[t^{-2\alpha-2}-2t^{-2\alpha-2}(1-t)^\alpha + t^{-2\alpha-2}(1-t)^{2\alpha}  \right]dt\\
&=\int_0^1t^{-2\alpha-2}dt+\int_0^1-2t^{-2\alpha-2}(1-t)^\alpha dt + \int_0^1t^{-2\alpha-2}(1-t)^{2\alpha}dt \\
&=I_1+I_2+I_3
\end{split}
\end{equation}
\begin{equation}
\begin{split}
I_1&=\left.\frac{t^{-2\alpha-1}}{-2\alpha-1}\right|_0^1=\infty\\
I_2&=-2B(-2\alpha-1,\alpha+1)=-2\frac{\Gamma(-2\alpha-1)\Gamma(\alpha+1)}{\Gamma(-\alpha)}\\
I_3&=B(-2\alpha-1,2\alpha+1)=\frac{\Gamma(-2\alpha-1)\Gamma(2\alpha+1)}{\Gamma(0)}=0\\
\end{split}
\end{equation}
The above derivations are not correct since beta function $B(w,v)$ requires $w,v>0$ and $I_1$ is divergent. What may be the problem?
Thank you

REPLY [3 votes]: Real Variable Approach
$$
\begin{align}
&\int_1^\infty(u^\alpha-(u-1)^\alpha)^2\,\mathrm{d}u\\
&=\lim_{\lambda\to\infty}\int_0^{\lambda-1}((u+1)^\alpha-u^\alpha)^2\,\mathrm{d}u\tag{1a}\\
&=\lim_{\lambda\to\infty}\left[\int_0^{\lambda-1}(u+1)^{2\alpha}\,\mathrm{d}u-2\int_0^{\lambda-1}(u+1)^\alpha u^\alpha\,\mathrm{d}u+\int_0^{\lambda-1}u^{2\alpha}\,\mathrm{d}u\right]\tag{1b}\\
&=\lim_{\lambda\to\infty}\small\left[\frac{\lambda^{2\alpha+1}-1}{2\alpha+1}+\frac{(\lambda-1)^{2\alpha+1}}{2\alpha+1}-2\int_0^{\lambda-1}(u+1)^\alpha u^\alpha\,\mathrm{d}u\right]\tag{1c}\\
&=\lim_{\lambda\to\infty}\small\left[\frac{\lambda^{2\alpha+1}-1}{2\alpha+1}+\frac{(\lambda-1)^{2\alpha+1}}{2\alpha+1}-\frac{2(\lambda-1)^{\alpha+1}\lambda^\alpha}{2\alpha+1}-\frac{2\alpha}{2\alpha+1}\int_0^{\lambda-1}(u+1)^{\alpha-1}u^\alpha\,\mathrm{d}u\right]\tag{1d}\\
&=\lim_{\lambda\to\infty}\small\left[\frac{\lambda^{2\alpha+1}-1}{2\alpha+1}+\frac{(\lambda-1)^{2\alpha+1}}{2\alpha+1}-\frac{2(\lambda-1)^{\alpha+1}\lambda^\alpha}{2\alpha+1}-\frac{(\lambda-1)^{\alpha+1}\lambda^{\alpha-1}}{2\alpha+1}\right]\\
&-\lim_{\lambda\to\infty}\frac{\alpha-1}{2\alpha+1}\int_0^{\lambda-1}(u+1)^{\alpha-2} u^{\alpha}\,\mathrm{d}u\tag{1e}\\[6pt]
&=\frac{1-\alpha}{2\alpha+1}\,\mathrm{B}(1+\alpha,1-2\alpha)-\frac1{2\alpha+1}\tag{1f}\\[6pt]
&=-2\,\mathrm{B}(1+\alpha,-1-2\alpha)-\frac1{2\alpha+1}\tag{1g}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $u\mapsto u+1$ and convert improper integral to a limit
$\text{(1b)}$: expand the square
$\text{(1c)}$: evaluate the left and right integrals of $\text{(1b)}$
$\text{(1d)}$: apply $(2)$ with $\beta=\alpha$
$\text{(1e)}$: apply $(2)$ with $\beta=\alpha-1$
$\text{(1f)}$: evaluate the limits using $(3)$
$\text{(1g)}$: use $\mathrm{B}(1+\alpha,1-2\alpha)=\frac{-2\alpha}{1-\alpha}\frac{-1-2\alpha}{-\alpha}\,\mathrm{B}(1+\alpha,-1-2\alpha)$

Integration by Parts to Reduce the Exponent of $\boldsymbol{(u+1)}$
$$
\begin{align}
&\int_0^{\lambda-1}(u+1)^\beta u^\alpha\,\mathrm{d}u\\
&=\frac1{a+1}\int_0^{\lambda-1}(u+1)^\beta\,\mathrm{d}u^{\alpha+1}\tag{2a}\\
&=\frac{(\lambda-1)^{\alpha+1}\lambda^\beta}{\alpha+1}-\frac\beta{\alpha+1}\int_0^{\lambda-1}(u+1)^{\beta-1}u^{\alpha+1}\,\mathrm{d}u\tag{2b}\\
&=\frac{(\lambda-1)^{\alpha+1}\lambda^\beta}{\alpha+1}-\frac\beta{\alpha+1}\left[\int_0^{\lambda-1}(u+1)^\beta u^\alpha\,\mathrm{d}u-\int_0^{\lambda-1}(u+1)^{\beta-1}u^\alpha\,\mathrm{d}u\right]\tag{2c}\\
&=\frac{(\lambda-1)^{\alpha+1}\lambda^\beta}{\alpha+\beta+1}+\frac\beta{\alpha+\beta+1}\int_0^{\lambda-1}(u+1)^{\beta-1}u^\alpha\,\mathrm{d}u\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: prepare to integrate by parts
$\text{(2b)}$: integrate by parts
$\text{(2c)}$: $u=(u+1)-1$
$\text{(2d)}$: add $\frac\beta{\alpha+1}\int_0^{\lambda-1}(u+1)^\beta u^\alpha\,\mathrm{d}u$ and multiply by $\frac{\alpha+1}{\alpha+\beta+1}$

Evaluation of a Limit From $\bf{(1e)}$
$$
\begin{align}
&\hphantom{=}\lambda^{2\alpha+1}+(\lambda-1)^{2\alpha+1}-2(\lambda-1)^{\alpha+1}\lambda^\alpha-(\lambda-1)^{\alpha+1}\lambda^{\alpha-1}\\
&=\lambda^{2\alpha}+\color{#C00000}{\lambda^{2\alpha}(\lambda-1)+(\lambda-1)^{2\alpha+1}-2(\lambda-1)^{\alpha+1}\lambda^\alpha}-(\lambda-1)^{\alpha+1}\lambda^{\alpha-1}\tag{3a}\\
&=\color{#00A000}{\lambda^{2\alpha}}+\color{#C00000}{(\lambda-1)\left(\lambda^\alpha-(\lambda-1)^\alpha\right)^2}\color{#00A000}{-(\lambda-1)^{\alpha+1}\lambda^{\alpha-1}}\tag{3b}\\
&=(\lambda-1)\left(\lambda^\alpha-(\lambda-1)^\alpha\right)^2+\color{#00A000}{\left(\lambda^{\alpha+1}-(\lambda-1)^{\alpha+1}\right)\lambda^{\alpha-1}}\tag{3c}\\
&\sim\lambda\left(\alpha\lambda^{\alpha-1}\right)^2+\left((\alpha+1)\lambda^\alpha\right)\lambda^{\alpha-1}\tag{3d}\\
&=(\alpha^2+\alpha+1)\lambda^{2\alpha-1}\tag{3e}\\
&\to0\tag{3f}
\end{align}
$$
Explanation:
$\text{(3a)}$: $\lambda^{2\alpha+1}=\lambda^{2\alpha}(1+(\lambda-1))=\lambda^{2\alpha}+\lambda^{2\alpha}(\lambda-1)$
$\text{(3b)}$: collect red terms
$\text{(3c)}$: collect green terms
$\text{(3d)}$: mean value theorem
$\text{(3e)}$: collect terms
$\text{(3f)}$: $2\alpha-1\lt0$