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Apparently, the value of this integral is:

$$I=\int_0^{\pi/2} \int_0^{\pi/2} \frac{\sin x-\sin y}{x-y} dx ~dy=2 \left(\text{Si}\left(\frac{\pi}{2} \right)+\text{Ci}\left(\frac{\pi}{2}\right)-\ln \left( \frac{\pi}{2} \right)-\gamma \right)$$

This is my attempt:

$$I=2\int_0^{\pi/2} \int_0^x \frac{\sin x-\sin y}{x-y} dx dy=$$

$$=2\int_0^{\pi/2} \int_0^x \frac{\sin ( \frac{x-y}{2})\cos ( \frac{x+y}{2})}{\frac{x-y}{2}} dx dy$$

Introducing new variables:

$$v=\frac{x+y}{2}, \qquad u=\frac{x-y}{2}$$

$$x=v+u, \qquad y=v-u$$

$$dx~dy=2~du~dv$$

We rewrite the integral:

$$I=4 \int_0^{\pi/4} \int_0^v \frac{\sin u \cos v}{u} du~ dv+4 \int_{\pi/4}^{\pi/2} \int_0^{\pi/2-v} \frac{\sin u \cos v}{u} du~ dv$$

Now we have:

$$I=4 \int_0^{\pi/4} \text{Si}\left(v \right) \cos v~ dv+4 \int_{\pi/4}^{\pi/2} \text{Si}\left(\frac{\pi}{2}-v \right) \cos v~ dv=$$

$$=4 \int_0^{\pi/4} \text{Si}\left(v \right) (\cos v+\sin v)~ dv$$

Is this correct so far?

We can check with Wolfram Alpha that the value of this last integral is as stated above.

How to proceed further?

Is there a better way?

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  • $\begingroup$ The answer has $\gamma$ and $-\ln(\pi/2)$, which makes me think it was derived by integrating the series expansions for $\sin(x)/(x-y)$ and $\sin(y)/(x-y)$ and then "recognizing" the series for Si and Ci. Someone, somewhere has integrated $1/x$. $\endgroup$ – B. Goddard Aug 23 '16 at 14:14
  • $\begingroup$ @B.Goddard, see my edit. I obtained the answer from WA for my last integral, not for the initial one. But thank you for the hint $\endgroup$ – Yuriy S Aug 23 '16 at 14:14
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I think the easiest way is to notice that the integral can be written as $\mathcal I(1)$, where $$\mathcal I(t):=\int_0^{\frac\pi 2}\int_0^{\frac\pi 2}\frac{2\sin\frac{t(x-y)}{2}\cos\frac{x+y}{2}\;dxdy}{x-y}.$$ Let us cancel the denominator by differentiating w.r.t. $t$, $$\mathcal I'(t)=\int_0^{\frac\pi 2}\int_0^{\frac\pi 2}\cos\frac{t(x-y)}{2}\cos\frac{x+y}{2}\;dxdy=\frac{4(\sqrt2\,\cos\frac{\pi t}{4}-1)}{1-t^2}.$$ Now integrating back, we will have $$\mathcal I(1)=\int_0^1\mathcal I'(t)dt=\int_0^1\frac{4(\sqrt2\,\cos\frac{\pi t}{4}-1)}{1-t^2}dt,$$ which is clearly expressible in terms of $\mathrm{Ci}$, $\mathrm{Si}$.

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