Computing a specific integral and expressing it in terms of the Gamma function

Question

Let $0< \alpha < \frac12, t_1>0, t_2 >0$. How to compute $$ \int_0^\infty \left( (x+t_1)^\alpha - x^\alpha \right) \left( (x+t_2)^{\alpha} - x^\alpha \right) \mathrm{d}x,$$ when $t_1 \not=t_2$? For $t_1= t_2\equiv t$ it is quite easy: $$ \int_0^\infty \left( (x+t)^\alpha - x^\alpha \right)^2 \mathrm{d}x = t^{2\alpha+1} \int_0^\infty \left( (\zeta +1)^\alpha - \zeta^\alpha \right)^2 \mathrm{d}\zeta \\= t^{2\alpha+1} \left[-\frac{1}{2\alpha+1}- \frac{ 2^{-2\alpha -1} \Gamma(-\alpha - \frac12) \Gamma(\alpha +1) }{\sqrt{\pi}} \right]$$ using the substitution $\zeta = \frac{x}{t} \Leftrightarrow \mathrm{d}{x} = t \mathrm{d}\zeta$.

But is there a way to express it in some form like this for $t_1 \not=t_2$?

Thx in advance!

Answer 1

So in fact w.o.l.g. we can choose $t_2=1$ and then calculate the antiderivative of each of the four summands of $$\left( (x+t_1)^\alpha - x^\alpha \right) \left( (x+t_2)^{\alpha} - x^\alpha \right)$$ and evaluate them at 0 and $\infty$, seeing which are asymptotically equal.

Answer 2

If you expand the product (as @cptflint already wrote), you need to compute four antiderivatives but not integrals $$I_1=\int x^{2\alpha}\,dx=\frac{x^{2 \alpha+1}}{2 \alpha+1}$$ $$I_2=\int x^{\alpha}(x+t_1)^\alpha\,dx=\frac{x^{ \alpha+1}}{ \alpha+1}\, t_1^\alpha\, _2F_1\left(-a,a+1;a+2;-\frac{x}{t_1}\right)$$ $$I_3=\int x^{\alpha}(x+t_2)^\alpha\,dx=\frac{x^{ \alpha+1}}{ \alpha+1}\, t_2^\alpha\, _2F_1\left(-a,a+1;a+2;-\frac{x}{t_2}\right)$$ $$I_4=\int (x+t_1)^\alpha(x+t_2)^\alpha\,dx=\frac{(x+t_1)^{ \alpha+1}}{ \alpha+1}\, (t_2-t_1)^\alpha\, _2F_1\left(-a,a+1;a+2;-\frac{x+t_1}{t_2-t_1}\right)$$ where the Gaussian hypergeometric functions could be replaced by incomplete beta functions.

Computing $$I=(I_1+I_4)-(I_2+I_3)$$ does not make any problem at the lower bound but it could be a bit more delicate when $x\to \infty$.

Edit

Assuming $t_1< t_2$, given by Mathematica

$$I=\int_0^y \left( (x+t_1)^\alpha - x^\alpha \right) \left( (x+t_2)^{\alpha} - x^\alpha \right) \,dx=$$ $$I=\frac{y^{2 \alpha+1}}{2 \alpha+1}+e^{-i \pi \alpha} \Big[\cdots \Big]$$ with $$\Big[\cdots \Big]=t_1^{2 \alpha+1} B_{-\frac{y}{t_1}}(\alpha+1,\alpha+1)+t_2^{2 \alpha+1} B_{-\frac{y}{t_2}}(\alpha+1,\alpha+1)+$$ $$(t_2-t_1)^{2 \alpha+1} \left(B_{\frac{t_1}{t_1-t_2}}(\alpha+1,\alpha+1)-B_{\frac{t_1+y}{t_1-t_2}}(\alpha+1,\alpha+1) \right)$$