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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 Sorry, that comment I made above was before I edited my post - I initially thought a simple integration by parts would solve the problem, but it didn't. You need to change the limits of integration.2012-05-21

1 Answers 1

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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. $

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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](http://www.youtube.com/watch?v=NETmfwOAKpQ) and [2](http://mathinsight.org/double_integral_change_order_integration_examples).2012-04-18