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Calculate: $$\int_0^\pi \int_x^\pi \frac{\sin y}{y} dydx$$ How to calculate that? This x is terribly confusing for me. I do not know how to deal with it properly.

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    $\begingroup$ When you do the inner integral $\int_x^\pi\frac{\sin y}{y}dy$, you treat $x$ just as though it were any other constant. Can you calculate $\int_a^\pi\frac{\sin y}{y}dy$? $\endgroup$
    – Arthur
    Aug 20, 2015 at 12:05
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    $\begingroup$ @Arthur : but $\int_x^\pi \frac{\sin(y)}{y} dy$ has no simple expression, so it won't help. $\endgroup$
    – Tryss
    Aug 20, 2015 at 12:10
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    $\begingroup$ @Tryss While you might be right, as I interpret the question, the OP is asking how to handle that $x$. I told him how to handle that $x$. How to handle the integral itself is a completely different question. $\endgroup$
    – Arthur
    Aug 20, 2015 at 12:41

3 Answers 3

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You need to invert both integrals with Fubini theorem (the function is positive, so no problem here):

$$I = \int_0^\pi \int_x^\pi \frac{\sin(y)}{y} dy dx = \int_D \frac{\sin(y)}{y} dx dy$$

Where $D = \{ (x,y) : 0\leq x \leq \pi\text{ and } x \leq y \leq \pi \}$

But you can also rewrite $D$ as $D = \{ (x,y) : 0\leq x \leq y \leq \pi\}$

Or, another form, $D = \{ (x,y) : 0\leq y \leq \pi \text{ and } 0 \leq x \leq y\}$

So we have

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

$$ = \int_0^\pi \frac{\sin(y)}{y} \left( \int_0^ydx \right) dy$$

$$ = \int_0^\pi \frac{\sin(y)}{y} y dy$$

$$ = \int_0^\pi \sin(y) dy$$

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  • $\begingroup$ could you explain the reason for this boundary change i.e. how to compute those boundaries? $\endgroup$ Aug 20, 2015 at 12:12
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    $\begingroup$ @mkropkowski : ok, I'll add this $\endgroup$
    – Tryss
    Aug 20, 2015 at 12:15
  • $\begingroup$ @Tryss hey, this may be a bit late. But isn't siny/y discontinuous at y = 0 ? $\endgroup$ Apr 5, 2020 at 5:02
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You're looking to integrate $\dfrac{\sin y}{y}$ over this area: enter image description here

where $x\le y\le \pi$ and then $0 \le x \le \pi$. But notice that if you swap the axes:

enter image description here

You'll have the same area but seen from a different perspective where $0 \le x\le y$ and $0\le y \le \pi$, which graphically justifies the following application of Fubini's Theorem:

$$\int_0^\pi \int_x^\pi \frac{\sin y}{y} dydx=\int_0^\pi \int _0^y \frac{\sin y}{y} dxdy=\int_0^\pi \sin y dy=2 $$

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Let $f: (x,y) \mapsto y^{-1}\sin y$ for all $(x,y) \in \mathbb{R}^{2}$ such that $y \neq 0$; let $S := \{ (x,y) \in \mathbb{R}^{2} \mid 0 \leq x \leq \pi, x \leq y \leq \pi \}$; and let $\int_{S} f$ exist. Then by Fubini's theorem we have $$ \int_{S}f = \int_{0}^{\pi}\int_{x}^{\pi} \frac{\sin y}{y} dy dx = \int_{0}^{\pi}\int_{0}^{y} \frac{\sin y}{y} dx dy = \int_{0}^{\pi}\sin y dy = -\cos y \big|_{0}^{\pi} = 2. $$

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    $\begingroup$ Ha, observe that $S = \{ (x,y) \in \mathbb{R}^{2} \mid 0 \leq y \leq \pi, 0 \leq x \leq y \}$. $\endgroup$
    – Megadeth
    Aug 20, 2015 at 12:16
  • $\begingroup$ Also please note that it is important to assume the existence of the double integral. Simply being integrable iteratively does not ensure the existence of a double integral. $\endgroup$
    – Megadeth
    Aug 20, 2015 at 12:19
  • $\begingroup$ I thought it does if the integrand is non-negative? $\endgroup$
    – user21820
    Aug 20, 2015 at 16:33

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