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Sub-problem, $\int^{\pi/2}_y \frac{\sin x}{x} dx $, emerging on the page 941 p4b here which asks us to find :

$$\int^{\pi/2}_0 \left( \int^{\pi/2}_y \frac{\sin x}{x} dx \right) dy .$$

My instructor once showed me some nice deduction for the thing inside the integral, I think it was with exponential functions or something like that. Now the upper border can be handled with the 1st fundamental rule of calculus but I cannot remember how to proceed with the integral, how to proceed here?

Ragib suggested me to draw but failure, ideas how to fix it? (some oddity with $x_{0}$)

enter image description here

Some observations

Let $f(x,y) := \int_{\pi/2}^y \frac{\sin(x)}{x} dx$.

  1. if $y>\frac{\pi}{2}$, then we have no division by zero -case.

  2. If $y<\frac{\pi}{2}$ then we may have an indefinite case a bit more problematic situation, particularly when $y<0$.

  3. If an upper bound for $\sin(x)$ -- look they start from origin and the linear function is below $\sin(x)$ until point $p_{1}$, then there may be some theorem to use (perhaps some Cauchy-something, researching).

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hhh
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1 Answers1

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Interchange the order of integration:

$$ \int^{\pi/2}_0 \int^{\pi/2}_y \frac{\sin x}{x} dx dy = \int^{\pi/2}_0 \int^{x}_0 \frac{\sin x}{x} dy dx = \int^{\pi/2}_0 \sin x dx = 1. $$

Ragib Zaman
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  • @J.M. Yea I stuffed up, fixed now! – Ragib Zaman Apr 18 '12 at 13:30
  • something odd with borders, could you add some details?

    $\int^{\pi/2}_0 \int^{\pi/2}_y \frac{\sin x}{x} dx dy = \int^{\pi/2}_0 \int^{x}_0 \frac{\sin x}{x} dy dx$?

     – hhh Apr 18 '12 at 13:31
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    @hhh Have you learned how to adjust the limits of double integrals after you change their order? Drawing a picture can help you see the new limits. In general, $$ \int^{a}_0 \int^a_y f(x,y) dx dy = \int^a_0 \int^x_0 f(x,y) dx dy. $$ – Ragib Zaman Apr 18 '12 at 13:36
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    Making a drawing of the region is always helpful. If you're unable or unwilling to draw the region, you can use Iverson brackets instead: $[0\leq y\leq \pi/2][y\leq x\leq \pi/2]=[0\leq y\leq x\leq \pi/2]=[0\leq y\leq x][0\leq x\leq \pi/2]$ – J. M. ain't a mathematician Apr 18 '12 at 13:55
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    @hhh That's not quite the idea on how to do it. Here are two helpful resources that teach you: 1 and 2. – Ragib Zaman Apr 18 '12 at 14:06