Extending the result $\int_{0}^{\infty} \left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} \right) \, dx = \frac{4}{\pi} $

While generalizing this result, I succeeded in proving that for $\alpha > 0$, $\beta < 1$ and $1 < 2\alpha + \beta < 3$, we have

\begin{align*} &\int_{0}^{\infty} \left[ \left( \int_{x}^{\infty} \frac{\cos t}{t^{\alpha}} \, dt \right)^{2} + \left( \int_{x}^{\infty} \frac{\sin t}{t^{\alpha}} \, dt \right)^{2} \right] \, \frac{dx}{x^{\beta}} \\ & \hspace{10em} = \frac{\pi}{1-\beta} \frac{\Gamma(2-\alpha-\beta)}{\Gamma(\alpha)} \csc \left( \pi \alpha + \frac{\pi (\beta-1)}{2} \right). \end{align*}

My question is

1. Is this a known result?
2. My ultimate goal is to examine whether the integral $$ I(\alpha, \beta) := \int_{0}^{\infty} \left[ \left( \int_{x}^{\infty} \frac{\cos t}{t^{\alpha}} \, dt \right)^{2} + \left( \int_{x}^{\infty} \frac{\sin t}{t^{\alpha}} \, dt \right)^{2} \right]^{2} \, \frac{dx}{x^{\beta}} $$ has closed from or not for general $\alpha$ and $\beta$. I know that $$ I \left(\tfrac{1}{2}, 0 \right) = 2\pi (\log 4 - 1) \qquad \text{and} \qquad I \left(1, 0 \right) = \frac{2\pi^{3}}{3}, \tag{1} $$ but I know nothing for the other cases. (Here, the former identity in $(1)$ corresponds to the motivating problem linked above.) Is there any other known result concerning this integral?

A further inspection showed that

\begin{align*} \int_{0}^{\infty} \left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} \right)^{2} \, dx &= \frac{16}{\pi} \left( 1 - \frac{2\sqrt{2}}{\pi} \log \left( 1 + \sqrt{2} \right) \right) \\ &\approx 1.0516193625061961290 \cdots, \end{align*}

where

$$ C(x) = \int_{0}^{x} \cos \left( \tfrac{\pi t^2}{2} \right) \, dt \quad \text{and} \quad S(x) = \int_{0}^{x} \sin \left( \tfrac{\pi t^2}{2} \right) \, dt $$

are Fresnel integrals. Indeed, this corresponds to

$$ I \left( \tfrac{1}{2}, \tfrac{1}{2} \right) = \sqrt{\pi} \left( 4 \sqrt{2} \pi - 16 \log \left( 1 + \sqrt{2} \right) \right). $$

Note that major inverse symbolic calculators do not yield this result.

Solution 1:

Pardon My Progress...

First, the sum of squared integrals inside the brackets, which for convenience we shall denote by $f{\left(x;\alpha\right)}$, may be rewritten as a single double integral with a little algebra and trigonometry:

$$\begin{align} f{\left(x;\alpha\right)} &:=\left(\int_{x}^{\infty}\frac{\cos{t}}{t^\alpha}\,\mathrm{d}t\right)^2+\left(\int_{x}^{\infty}\frac{\sin{t}}{t^\alpha}\,\mathrm{d}t\right)^2\\ &=\small{\left(\int_{x}^{\infty}\frac{\cos{t_1}}{t_1^\alpha}\,\mathrm{d}t_{1}\right)\cdot\left(\int_{x}^{\infty}\frac{\cos{t_2}}{t_2^\alpha}\,\mathrm{d}t_{2}\right)+\left(\int_{x}^{\infty}\frac{\sin{t_1}}{t_1^\alpha}\,\mathrm{d}t_{1}\right)\cdot\left(\int_{x}^{\infty}\frac{\sin{t_2}}{t_2^\alpha}\,\mathrm{d}t_{2}\right)}\\ &=\int_{x}^{\infty}\int_{x}^{\infty}\left(\frac{\cos{t_1}}{t_1^\alpha}\cdot\frac{\cos{t_2}}{t_2^\alpha}\right)\,\mathrm{d}t_{1}\mathrm{d}t_{2}+\int_{x}^{\infty}\int_{x}^{\infty}\left(\frac{\sin{t_1}}{t_1^\alpha}\cdot\frac{\sin{t_2}}{t_2^\alpha}\right)\,\mathrm{d}t_{1}\mathrm{d}t_{2}\\ &=\int_{x}^{\infty}\int_{x}^{\infty}\left(\frac{\cos{\left(t_1\right)}\cos{\left(t_2\right)}+\sin{\left(t_1\right)}\sin{\left(t_2\right)}}{\left(t_{1}t_{2}\right)^\alpha}\right)\,\mathrm{d}t_{1}\mathrm{d}t_{2}\\ &=\int_{x}^{\infty}\int_{x}^{\infty}\frac{\cos{\left(t_1-t_{2}\right)}}{\left(t_{1}t_{2}\right)^\alpha}\,\mathrm{d}t_{1}\mathrm{d}t_{2}.\\ \end{align}$$

Next, we apply a sequence of two two-variable transformations to put the integral into a more tractable form. The first substitution is a simple scaling transformation, $(1)$$(t_{1},t_{2})=(xu_{1},xu_{2})$; the second substitution is the more complicated transformation, $(2)$ $(u_{1}-u_{2},u_{1}u_{2})=(w_{1},w_{2})$:

$$\begin{align} \mathcal{I}{\left(\alpha,\beta\right)} &=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[\left(\int_{x}^{\infty}\frac{\cos{t}}{t^\alpha}\,\mathrm{d}t\right)^2+\left(\int_{x}^{\infty}\frac{\sin{t}}{t^\alpha}\,\mathrm{d}t\right)^2\right]^2\\ &=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[f{\left(x;\alpha\right)}\right]^2\\ &=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[\int_{x}^{\infty}\int_{x}^{\infty}\frac{\cos{\left(t_1-t_{2}\right)}}{\left(t_{1}t_{2}\right)^\alpha}\,\mathrm{d}t_{1}\mathrm{d}t_{2}\right]^2\\ &=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[\int_{1}^{\infty}\int_{1}^{\infty}\frac{\cos{\left[x\left(u_1-u_{2}\right)\right]}}{\left(x^2u_{1}u_{2}\right)^\alpha}\,x^2\,\mathrm{d}u_{1}\mathrm{d}u_{2}\right]^2\tag{1}\\ &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{1}^{\infty}\int_{1}^{\infty}\frac{\cos{\left[x\left(u_1-u_{2}\right)\right]}}{\left(u_{1}u_{2}\right)^\alpha}\,\mathrm{d}u_{1}\mathrm{d}u_{2}\right]^2\\ &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{1}^{\infty}\mathrm{d}w_{2}\int_{1-w_{2}}^{w_{2}-1}\mathrm{d}w_{1}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{w_{1}^2+4w_{2}}}\right]^2\tag{2}\\ &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{-\infty}^{\infty}\mathrm{d}w_{1}\int_{1+|w_{1}|}^{\infty}\mathrm{d}w_{2}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{w_{1}^2+4w_{2}}}\right]^2\\ &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[2\int_{0}^{\infty}\mathrm{d}w_{1}\int_{1+w_{1}}^{\infty}\mathrm{d}w_{2}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{w_{1}^2+4w_{2}}}\right]^2\\ &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}w_{1}\int_{1+w_{1}}^{\infty}\mathrm{d}w_{2}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{\frac14w_{1}^2+w_{2}}}\right]^2\\ &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}w_{1}\,\cos{\left(x\,w_{1}\right)}\int_{1+w_{1}}^{\infty}\,\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}}\right]^2.\\ \end{align}$$

To perform the integration with respect to $w_{2}$, I appeal to Gradshteyn 3.197(2): under the conditions $|\arg{\frac{u}{\beta}}|<\pi\lor |\frac{\beta}{u}|<1$, and $0<\Re{\left(\mu\right)}<\Re{\left(\lambda-\nu\right)}$, we have the result,

$$\int_{u}^{\infty}x^{-\lambda}(x+\beta)^{\nu}(x-u)^{\mu-1}\,\mathrm{d}x=u^{-\lambda}(\beta+u)^{\mu+\nu}\operatorname{B}{\left(\lambda-\mu-\nu,\mu\right)}\times{_2F_1}{\left(\lambda,\mu;\lambda-\mu;-\frac{\beta}{u}\right)}.$$

Let $\mu=1$, $\nu=-\frac12$, $\lambda=\alpha$, $u=1+w_{1}$, and $\beta=\frac14w_{1}^2$. Then for $\frac12<\Re{\left(\alpha\right)}<1\land w_{1}>0$,

$$\begin{align} \int_{1+w_{1}}^{\infty}\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}} &=\small{\left(w_{1}+1\right)^{-\alpha}\sqrt{\frac14w_{1}^2+w_{1}+1}\,\operatorname{B}{\left(\alpha-\frac12,1\right)}\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{\frac14w_{1}^2}{1+w_{1}}\right)}}\\ &=\left(w_{1}+1\right)^{-\alpha}\left(\frac{w_{1}+2}{2}\right)\,\left(\frac{2}{2\alpha-1}\right)\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}\\ &=\frac{\left(w_{1}+2\right)\left(w_{1}+1\right)^{-\alpha}}{2\alpha-1}\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}.\\ \end{align}$$

The hypergeometric function above quite conveniently reduces to a rational function for the specified combination of parameters: for $\alpha\neq1$,

$${_2F_1}{\left(\alpha,1;\alpha-1;-z\right)}=\frac{\alpha\,z+\alpha-2z-1}{(\alpha-1)(z+1)^2},\\ \implies {_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}=\frac{4\left(w_{1}+1\right)\left[\alpha(w_{1}+2)^2-2(w_{1}+1)^2-2\right]}{(\alpha-1)(w_{1}+2)^4}.\\$$

Thus, for $\frac12<\Re{\left(\alpha\right)}<1\land w_{1}>0$,

$$\begin{align} \int_{1+w_{1}}^{\infty}\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}} &=\frac{\left(w_{1}+2\right)\left(w_{1}+1\right)^{-\alpha}}{2\alpha-1}\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}\\ &=\frac{4\left(w_{1}+1\right)^{1-\alpha}\left[\alpha(w_{1}+2)^2-2(w_{1}+1)^2-2\right]}{(2\alpha-1)(\alpha-1)(w_{1}+2)^3},\\ \end{align}$$

and hence:

$$\begin{align} \mathcal{I}{\left(\alpha,\beta\right)} &=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}w_{1}\,\cos{\left(x\,w_{1}\right)}\int_{1+w_{1}}^{\infty}\,\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}}\right]^2\\ &=\small{\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta-4\left(1-\alpha\right)}}\left[\int_{0}^{\infty}\frac{4\left(w_{1}+1\right)^{1-\alpha}\left[\alpha(w_{1}+2)^2-2(w_{1}+1)^2-2\right]\cos{\left(xw_{1}\right)}}{(2\alpha-1)(\alpha-1)(w_{1}+2)^3}\,\mathrm{d}w_{1}\right]^2}\\ &=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta-4\left(1-\alpha\right)}}\left[\int_{0}^{\infty}\frac{4\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]\cos{\left(xy\right)}}{(2\alpha-1)(\alpha-1)(y+2)^3}\,\mathrm{d}y\right]^2\\ &=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta-4\left(1-\alpha\right)}}\left[\int_{0}^{\infty}\frac{\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]\cos{\left(xy\right)}}{(y+2)^3}\,\mathrm{d}y\right]^2},\\ &=:\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]^2,\\ \end{align}$$

where in the third line we've made the substitution $w_{1}=y$ for the sake of eliminating the dependency on subscripted variables, and where in the last line we've introduced the auxiliary function $g{\left(y;\alpha\right)}$ simply for the sake conveniently denoting the function,

$$\begin{align} g{\left(y;\alpha\right)} &:=\frac{\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]}{(y+2)^3}\\ &=\frac{\alpha\left(y+1\right)^{1-\alpha}}{y+2}-\frac{2\left(y+1\right)^{3-\alpha}}{(y+2)^3}-\frac{2\left(y+1\right)^{1-\alpha}}{(y+2)^3}.\\ \end{align}$$

Now if we repeat our initial trick of rewriting the square of an integral as a double integral, we can arrive at an expression for $\mathcal{I}{\left(\alpha,\beta\right)}$ as an ordinary triple integral:

$$\begin{align} \mathcal{I}{\left(\alpha,\beta\right)} &=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]^2\\ &=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]\cdot\left[\int_{0}^{\infty}g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\,\mathrm{d}z\right]}\\ &=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}\right]}\\ &=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}.\\ \end{align}$$

Change the order of integration so that the integration over $x$ is first:

$$\begin{align} \mathcal{I}{\left(\alpha,\beta\right)} &=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\ &=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\int_{0}^{\infty}\mathrm{d}x\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\ &=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\int_{0}^{\infty}\mathrm{d}x\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta-4\left(1-\alpha\right)}}.\\ \end{align}$$

Then the inner integral with respect to $x$ may be evaluated in closed form with an appeal to another result from Gradshteyn. Proposition 3.762(3) of Gradshteyn states that, given $a,b\in\mathbb{R}$ and $\mu\in\mathbb{C}$ such that $a>0,~b>0$, and $0<\Re{(\mu)}<1$, then the following improper integral has the closed form:

$$\int_{0}^{\infty}x^{\mu-1}\cos{\left(ax\right)}\cos{\left(bx\right)}\,\mathrm{d}x=\frac12\cos{\left(\frac{\mu\pi}{2}\right)}\,\Gamma{\left(\mu\right)}\,\left[\left(a+b\right)^{-\mu}+|a-b|^{-\mu}\right].$$

Setting $(a,b,\mu)\mapsto(y,z,4(1-\alpha)-\beta+1)$ in the above proposition yields the following corrollary: given $y,z\in\mathbb{R}$ and $p\in\mathbb{C}$ such that $y>0,~z>0$, and $-1<\Re{\left(4(1-\alpha)-\beta\right)}<0$, then the following improper integral has the closed form,

$$\small{\int_{0}^{\infty}\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta+4\alpha-4}}\mathrm{d}x=\frac12\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}\,\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right].}$$

Thus, we may reduce our integral representation of $\mathcal{I}{\left(\alpha,\beta\right)}$ to a single double integral:

$$\begin{align} \mathcal{I}{\left(\alpha,\beta\right)} &=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\int_{0}^{\infty}\mathrm{d}x\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\ &=\small{\frac{8\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]}\\ &=:\frac{8\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}}{(2\alpha-1)^2(\alpha-1)^2}\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}.\\ \end{align}$$

Update:

Now we'll focus on reducing the previously defined function $\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}$. First of all, by symmetry we can reduce the region of integration to one where the absolute value bars are no longer necessary inside the integrand, which will obviate some of tedium of evaluation:

$$\begin{align} \tilde{\mathcal{I}}{\left(\alpha,\beta\right)} &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{y}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &~~~~~ +\int_{0}^{\infty}\mathrm{d}z\int_{0}^{z}\mathrm{d}y\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(z;\alpha\right)}\,g{\left(y;\alpha\right)}\left[\left(z+y\right)^{\beta+4\alpha-5}+|z-y|^{\beta+4\alpha-5}\right]\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\ &=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+\left(y-z\right)^{\beta+4\alpha-5}\right].\\ \end{align}$$

Next, we rescale the interval of integration of the inner integral to the unit interval via the substitution $z=y\,\omega$:

$$\begin{align} \tilde{\mathcal{I}}{\left(\alpha,\beta\right)} &=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+\left(y-z\right)^{\beta+4\alpha-5}\right]\\ &=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{1}y\,\mathrm{d}\omega\,g{\left(y;\alpha\right)}\,g{\left(y\,\omega;\alpha\right)}\left[\left(y+y\,\omega\right)^{\beta+4\alpha-5}+\left(y-y\,\omega\right)^{\beta+4\alpha-5}\right]\\ &=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{1}y^{\beta+4\alpha-4}\,\mathrm{d}\omega\,g{\left(y;\alpha\right)}\,g{\left(y\,\omega;\alpha\right)}\left[\left(1+\omega\right)^{\beta+4\alpha-5}+\left(1-\omega\right)^{\beta+4\alpha-5}\right]\\ &=2\int_{0}^{\infty}\mathrm{d}y\,y^{\beta+4\alpha-4}\,g{\left(y;\alpha\right)}\int_{0}^{1}\mathrm{d}\omega\,g{\left(y\,\omega;\alpha\right)}\left[\left(1+\omega\right)^{\beta+4\alpha-5}+\left(1-\omega\right)^{\beta+4\alpha-5}\right].\\ \end{align}$$

At this stage, the inner integral over $\omega$ may be evaluated in terms of the Appell hypergeometric function, $F_{1}$.