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$$\int_{0}^{\pi}\int_{0}^{x}\frac{\sin y}{\pi-y}\,dy\,dx = \large{?}$$

For this question,I changed the order of limits and then tried the substitution for $(\pi-y)$ as $t$ and tried integration by parts but I couldn't proceed further as the answer was getting really complicated and this is a numerical answer type question

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Let $I$ be the integral given by

$$I=\int_0^\pi \int_0^x \frac{\sin(y)}{\pi-y}\,dy\,dx$$

ASIDE:

Since $\left|\frac{\sin(y)}{\pi-y}\right|\le 1$, $\iint\left|\frac{\sin(y)}{\pi-y}\right|\,d(x,y)<\infty$. Therefore, Fubini's Theorem guarantees that the double integral and both iterated integrals are equal.

Changing the order of integration reveals

$$\begin{align} I&=\int_0^\pi \int_y^\pi \frac{\sin(y)}{\pi-y}\,dx\,dy\\\\ &=\int_0^\pi \sin(y)\,dy\\\\ &=2 \end{align}$$

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  • $\begingroup$ can you explain how you changed the order of integration? $\endgroup$
    – PiGamma
    Commented Oct 1, 2016 at 15:18
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    $\begingroup$ Sure. The domain of integration is a triangular region with vertices at $(0,0)$, $(\pi,0)$, and $(\pi,\pi)$. So, this can be described as the region $x\in [0,\pi] \times y\in [0,x]$ OR $y\in [0,\pi] \times x\in [y,\pi]$. $\endgroup$
    – Mark Viola
    Commented Oct 1, 2016 at 15:22
  • $\begingroup$ I would advise @NAJMAASHRAF to take a look at Fubini's theorem! If it may help $\endgroup$
    – Enrico M.
    Commented Oct 1, 2016 at 15:30
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    $\begingroup$ @FourierTransform And since the integrand has a removable discontinuity at $\pi$, the interchange of integrals is legitimate. $\endgroup$
    – Mark Viola
    Commented Oct 1, 2016 at 15:47

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