# Double integral problem: $\int_0^\pi\int_x^\pi \frac{\sin y}{y} dy\, dx$

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.

• 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 $\int_a^\pi\frac{\sin y}{y}dy$? Aug 20, 2015 at 12:05
• @Arthur : but $\int_x^\pi \frac{\sin(y)}{y} dy$ has no simple expression, so it won't help. Aug 20, 2015 at 12:10
• @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. Aug 20, 2015 at 12:41

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

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

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

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

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

• could you explain the reason for this boundary change i.e. how to compute those boundaries? Aug 20, 2015 at 12:12
• @mkropkowski : ok, I'll add this Aug 20, 2015 at 12:15
• @Tryss hey, this may be a bit late. But isn't siny/y discontinuous at y = 0 ? Apr 5, 2020 at 5:02

You're looking to integrate $\dfrac{\sin y}{y}$ over this area:

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

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:

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

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 $$\int_{S}f = \int_{0}^{\pi}\int_{x}^{\pi} \frac{\sin y}{y} dy dx = \int_{0}^{\pi}\int_{0}^{y} \frac{\sin y}{y} dx dy = \int_{0}^{\pi}\sin y dy = -\cos y \big|_{0}^{\pi} = 2.$$

• Ha, observe that $S = \{ (x,y) \in \mathbb{R}^{2} \mid 0 \leq y \leq \pi, 0 \leq x \leq y \}$. Aug 20, 2015 at 12:16
• 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. Aug 20, 2015 at 12:19
• I thought it does if the integrand is non-negative? Aug 20, 2015 at 16:33