2

The integral is: $\int^{\pi}_0\int^{\pi}_x \frac{\sin(y)} {y} dy dx$. I don't know if in this case I can change the order of the integral, but if so, I would have to integrate $\frac{\sin(x)} {x}$ in any case, so I don't know how to solve this integral.

zoli
  • 20,452
Josemi
  • 167
  • See here: http://math.stackexchange.com/questions/163305/what-is-the-integral-of-function-fx-sin-x-x – zoli Feb 04 '17 at 12:08
  • But even tough I know that the integral of $\frac {sin(x)} {x}$ is Si(x) I would have to integrate Si(x). – Josemi Feb 04 '17 at 12:33

2 Answers2

5

Your thinking that "I would have to integrate $\frac{sin(x)}{x}$ in any case" is wrong. Changing the order of integration gives $\int_0^\pi\int_0^y \frac{sin(y)}{y}dxdy= \int_0^\pi \left[\frac{sin(y)}{y}x\right]_{x=0}^y dy= \int_0^\pi \frac{sin(y)}{y}(y- 0)dy= \int_0^\pi sin(y)dy$.

user247327
  • 18,710
  • And why is that the change? My teacher haven't explain it to me, I would appreciate if you do it. – Josemi Feb 04 '17 at 18:02
  • Look at the region of integration: in the original integral, x goes from 0 to $\pi$ and, for each x, y goes from x to $\pi$. If x and y both went from 0 to $\pi$, that region would be the square with opposite vertices (0, 0) and $(\pi, \pi)$. y= x is the line connecting the two vertices and the condition that y> x restricts us to the triangular region above that line. Now, we can see that this triangle can be described as "y between 0 and $\pi$ and, for each y, x between 0 and y". That gives the limits of integration $\int_0^\pi\int_0^y dxdy$. – user247327 Feb 05 '17 at 13:44
  • Thanks, now I can do this type of integrals – Josemi Feb 06 '17 at 15:28
0

$$\int^{\pi}_0\int^{\pi}_x \frac{\sin(y)} {y} \ dy \ dx=-\int^{\pi}_0\int_{\pi}^x \frac{\sin(y)} {y} \ dy \ dx.$$

Now, take

$$\int^{\pi}_01\times\int_{\pi}^x \frac{\sin(y)} {y} \ dy \ dx$$

and integrate by part with $v'=1$ and $u=\int_{\pi}^x \frac{\sin(y)} {y} \ dy.$

We have then

$$\left[x\int_{\pi}^x\ \frac{\sin(y)} {y}dy\right]_0^{\pi}-\int_0^{\pi}x \frac{d}{d x}\int_{\pi}^x\frac{\sin(y)} {y}\ dy\ dx=$$ $$-\int_0^{\pi}\sin(x) \ dx=-2.$$

Because

$$\left[x\int_{\pi}^x\ \frac{\sin(y)} {y}dy\right]_0^{\pi}=\left[x(\operatorname {Si}(x)-\operatorname {Si}(\pi))\right]_0^{\pi}=0.$$

That is,

$$\int^{\pi}_0\int^{\pi}_x \frac{\sin(y)} {y} \ dy \ dx=2.$$

zoli
  • 20,452