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This is a follow-up question to A difficult 2d trigonometric integral. Unfortunately, I had a mistake in my calculations and I need a different (yet similar) seemingly simple integral solved:

$$int_{a}^{b}int_{a}^{y}frac{sin(x-y)}{xy}mathop{mathrm{d}x}mathop{mathrm{d}y}$$

For some $0<a<b$. The same methods as before can not be applied here, and I've been sitting for hours trying to solve this. Solutions (also using $mathop{mathrm{Si}}$ & $mathop{mathrm{Ci}}$ functions) would be very-very-very appreciated.

Thanks!

First, use the identity $sin (x-y)=sin x cos y - cos x sin y$, and the sum rule to get $$int_a^bleft[int_a^yfrac{sin xcos y}{xy}mathrm{d}x-int_a^yfrac{cos xsin y}{xy} mathrm{d}x right]mathrm{d}y$$
Pull out the constants and you get $$int_a^bleft[ frac{cos{y}}{y}int_a^yfrac{sin x}{x}mathrm{d}x-frac{sin y}{y}int_a^yfrac{cos x}{x}mathrm{d}x right]mathrm{d}y$$

This simplifies to $$int_a^bleft[frac{cos y}{y}Bigr[mathrm{Si}(x)Bigr]_a^y-frac{sin y}{y}Bigr[mathrm{Ci}(x)Bigr]_a^yright]mathrm{d}y$$

$$int_a^bleft[frac{cos y}{y}Bigr(mathrm{Si}(y)-mathrm{Si}(a)Bigr)-frac{sin y}{y}Bigr(mathrm{Ci}(y)-mathrm{Ci}(a)Bigr)right]mathrm{d}y$$

$$int_a^bleft[frac{mathrm{Si}(y)cos y}{y}-frac{mathrm{Si}(a)cos y}{y}-frac{mathrm{Ci}(y)sin y}{y}+frac{mathrm{Ci}(a)sin y}{y}right]mathrm{d}y$$

$$int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y-int_a^bfrac{mathrm{Si}(a)cos y}{y}mathrm{d}y-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y+int_a^bfrac{mathrm{Ci}(a)sin y}{y}mathrm{d}y$$

$mathrm{Si}(a)$ and $mathrm{Ci}(a)$ are constants, so they come out of their respective integrals, which further simplify. $$int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y-mathrm{Si}(a)int_a^bfrac{cos y}{y}mathrm{d}y-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y+mathrm{Ci}(a)int_a^bfrac{sin y}{y}mathrm{d}y$$

$$int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y-mathrm{Si}(a)Bigr[mathrm{Ci}(y)Bigr]_a^b-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y+mathrm{Ci}(a)Bigr[mathrm{Si}(y)Bigr]_a^b$$

$$mathrm{Si}(a)mathrm{Ci}(a)-mathrm{Si}(a)mathrm{Ci}(b)+mathrm{Si}(b)mathrm{Ci}(a)-mathrm{Si}(a)mathrm{Ci}(a)+int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y$$

$$mathrm{Si}(b)mathrm{Ci}(a)-mathrm{Si}(a)mathrm{Ci}(b)+int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y$$

Using integration by parts, we can transform $-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y$ into the other integral. $$-int_a^bfrac{mathrm{Ci}(y)sin y}{y}mathrm{d}y = -Bigr[mathrm{Si}(y)mathrm{Ci}(y)Bigr]_a^b+int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y$$

Substituting, we get $$mathrm{Si}(b)mathrm{Ci}(a)-mathrm{Si}(a)mathrm{Ci}(b)+mathrm{Si}(a)mathrm{Ci}(a)-mathrm{Si}(b)mathrm{Ci}(b)+2int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y$$

$$big(mathrm{Si}(a)+mathrm{Si}(b)big)big(mathrm{Ci}(a)-mathrm{Ci}(b)big)+2int_a^bfrac{mathrm{Si}(y)cos y}{y}$$

Now, to the best of my knowledge, that integral has no elementary solution. After laboring over it, I finally ran it through Maxima, which uses the Risch Algorithm, and it got nothing as well. However, we can expand the sine integral as a convergent infinite series (source: [Wikipedia][1]), and then solve the integral from there. $$mathrm{Si}(y)=sum_{n=0}^inftyfrac{(-1)^ny^{2n+1}}{(2n+1)(2n+1)!}$$
Substituting this in, we get $$2int_a^bfrac{mathrm{Si}(y)cos y}{y}mathrm{d}y=sum_{n=0}^inftyfrac{2(-1)^n}{(2n+1)(2n+1)!}int_a^bfrac{y^{2n-1}cos y}{y}mathrm{d}y$$
$$sum_{n=0}^inftyfrac{2(-1)^n}{(2n+1)(2n+1)!}int_a^b y^{2n}cos y,mathrm{d}y$$
That integral is solvable, and can be found to be
$$int_a^b y^{2n}cos y,mathrm{d}y=Bigr[-frac{1}{2}i^{2n+1}big[Gamma(2n+1,-ix)+(-1)^{2n}Gamma(2n+1,ix)big]Bigr]_a^b$$
Plugging this in we get the final answer.
$$-sum_{n=0}^inftyfrac{i^{4n+1}}{(2n+1)(2n+1)!}big[Gamma(2n+1,-ib)+(-1)^{2n}Gamma(2n+1,ib)-Gamma(2n+1,-ia)-(-1)^{2n}Gamma(2n+1,ia)big]$$
$$big(mathrm{Si}(a)+mathrm{Si}(b)big)big(mathrm{Ci}(a)-mathrm{Ci}(b)big)-sum_{n=0}^inftyfrac{i^{4n+1}}{(2n+1)(2n+1)!}big[Gamma(2n+1,-ib)+(-1)^{2n}Gamma(2n+1,ib)-Gamma(2n+1,-ia)-(-1)^{2n}Gamma(2n+1,ia)big]$$
I have it worked out without the incomplete Gamma function as a double sum. If you want me to write out that solution as well, leave a comment.

EDIT:
Here's the solution involving infinite sums.
Now, because n is discrete, I can describe the integral as an infinite sum. I solved the integrals for the cases $n = 0$, $n=1$, and $n=2$, and constructed the rule
$$int_a^bfrac{y^{2n-1}cos y}{y}mathrm{d}y=left[sum_{q=0}^{2n}(-1)^{q+1}frac{2n!}{q!}frac{mathrm{d}^{q}sin x}{mathrm{d}x^{q}}x^qright]_a^b$$
$$sum_{q=0}^{2n}(-1)^{q+1}frac{2n!}{q!}frac{mathrm{d}^{q}sin b}{mathrm{d}x^{q}}b^q-sum_{q=0}^{2n}(-1)^{q+1}frac{2n!}{q!}frac{mathrm{d}^{q}sin a}{mathrm{d}x^{q}}a^q$$
(I tested this for the case $n=3$, and it works fine). So, substituting this into the equation, we get
$$sum_{n=0}^inftyleft[frac{2(-1)^n}{(2n+1)(2n+1)!}left[sum_{q=0}^{2n}left[(-1)^{q+1}frac{2n!}{q!}frac{mathrm{d}^{q}sin x}{mathrm{d}x^{q}}x^qright]right]_a^bright]$$
$$big(mathrm{Si}(a)+mathrm{Si}(b)big)big(mathrm{Ci}(a)-mathrm{Ci}(b)big)+2sum_{n=0}^inftyleft[frac{2(-1)^n}{(2n+1)(2n+1)!}left[sum_{q=0}^{2n}left[(-1)^{q+1}frac{2n!}{q!}frac{mathrm{d}^{q}sin x}{mathrm{d}x^{q}}x^qright]right]_a^bright]$$