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I have been given this exercise: Calculate the double integral:

$$\iint_D\frac{\sin(y)}{y}dxdy$$ Where $D$ is the area enclosed by the lines: $y=2$, $y=1$, $y=x$, $2y=x$ (not $y = 2x$).

Visualising $D$ is easy. You can split D in two sub areas and get the bounds for the integrals. The problem I face is:

Let's split D in two sub areas, $D_1$ and $D_2$. $D_1$ is the left, upright triangle of $D$ and $D_2$ is the right, upside down one.

Then $D_1$ is defined by the lines $y=1$, $y=x$, and $x=2$.

You can express the area in a $y$-normal form as: $$\begin{align} 1 \le y \le 2\\ y \le x \le 2 \end{align}$$ then the integral can be written as $$ \begin{align} &\int_1^2\int_y^2\frac{\sin(y)}{y}dxdy \\ &=\int_1^2\frac{\sin(y)}{y}[x]^2_y \space dxdy \\ &=\int_1^2\frac{2\sin(y)}{y} - \sin(y)dy \\ &=2\int_1^2\frac{\sin(y)}{y}dy -\int_1^2 \sin(y)dy \\ \end{align}$$

The second integral is trivial, but in the first one is not. I have tried substituting, integrating by parts but to no avail. What am I doing wrong?

Any answer is really appreciated.

Giannis Koulouras
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    I think region should be: $y/2 < x < y; 1 < y < 2$. – GEdgar Apr 25 '22 at 21:38
  • Not sure if you set it up wrong, but if you are right, the first integral doesn't have an analytic solution, it's generally given its own function name, the Sine Integral. https://mathworld.wolfram.com/SineIntegral.html (Of which yours would be $Si(2)-Si(1)$ ) – Alan Apr 25 '22 at 21:40

2 Answers2

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Note that the integrand does not depend on $x$:

$$ \int\int_D \frac{\sin y}{y} dx dy = \int_1^2 \left( \int_{y/2}^y dx\right) \frac{\sin y}{y} dy = \int_1^2 \frac{y}{2}\frac{\sin y}{y} dy $$ $$ = \int_1^2 \frac{\sin y}{2} dy = \frac{\cos 1 - \cos 2}{2} $$

blamocur
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  • You are right, I needn't had split the area as it can be expressed just with y. I had mistyped the question ($2y=x$, not $y=2x$) so the inner integral bounds should be y≤x≤2y, but regardless, thank you very much. – Giannis Koulouras Apr 25 '22 at 22:29
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Under $x= \frac vu $, $y=v$, the Jacobian is $\frac v{u^2}$ and the integral becomes $$\iint_D\frac{\sin(y)}{y}dxdy=\int_1^2 \int_1^2 \frac{\sin v }{u^2}\ dv du=\sin\frac12\sin\frac32$$

Quanto
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