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I am trying to show that the Henstock-Kurzweil integral of sin(x)/x from zero to infinity is $\pi/2$. In order to do this, I wish to construct an explicit gauge and use Cousin's lemma to find a Perron partition so that the difference of $$ \left|\sum_{i=1}^n \frac{sin(\xi_i)}{\xi_i}(x_i-x_{i-1})\right|<\varepsilon. $$ My attempt is to say that our gauge, for any $\varepsilon$, is: $$ \delta_{\varepsilon}(\xi_i)=\xi_i\frac{i! 2^{i+1}}{4((2i+1)!)}+\frac{\varepsilon}{2^{k+1}} $$ This is a positive function, so by Cousins' lemma we have a Perron partition such that for every subinterval (start by finite endpoint), $x_i-x_{i-1} \leq 2\delta_{\varepsilon}(\xi_i)$. This means that our sum becomes $$ \left|\sum_{i=1}^n \frac{sin(\xi_i)}{\xi_i}(x_i-x_{i-1})\right|\leq \left|\sum_{i=1}^n \frac{sin(\xi_i)i!2^{i+1}}{2((2i+1)!)} +\sum_{i=1}^n \varepsilon2^{-k}-\frac{\pi}{2}\right|. $$ Note that $sin$ is bounded above by one, so now let n tend to infinity so we get that the expression above is strictly smaller than $$ \left|\sum_{i=1}^{\infty} \frac{i!2^{i+1}}{2((2i+1)!)} +\sum_{i=1}^{\infty}\varepsilon2^{-k}-\frac{\pi}{2}\right| $$ The first sum converges to $\pi/2$, the second to $\varepsilon$ and so we get that $$ \left|\sum_{i=1}^n \frac{sin(\xi_i)}{\xi_i}(x_i-x_{i-1})\right|<\varepsilon. $$ just as we wanted, and hence for every $\varepsilon$ we can find a gauge so that the difference and is small and we are done. But, this is wrong, is it not? Why is it wrong? I believe that it is something minor, in my sum inequalities.

Regards, edo.

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