Does the Absolute Value of the Sine Integral Diverge?

sparkster
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i know that the sine integral converges to pi/2. But what about the abs value of the sine integral. It seems to me that it would have value oo. But I'm having trouble coming up with a lower bound that diverges.
 
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Do you mean
the absolute value of Si(x).

\left|Si(x)\right|=:\left|\int_{0}^{x} \frac{\sin t}{t} \ dt \right|

or the abolute value of sinc(x)

\tilde{Si}(x)=:\int_{0}^{x} \left|\frac{\sin t}{t}\right| \ dt

Daniel.
 
dextercioby said:
Do you mean
the absolute value of Si(x).

\left|Si(x)\right|=:\left|\int_{0}^{x} \frac{\sin t}{t} \ dt \right|

or the abolute value of sinc(x)

\tilde{Si}(x)=:\int_{0}^{x} \left|\frac{\sin t}{t}\right| \ dt

Daniel.
The latter. Sorry for the confusion.
 
The graph is deceiving.My computer wouldn't compute the intagral.I don't know whether it's finite or not...

Daniel.
 
Break it up into intervals over the period of |sin(x)|

\int_{k*\pi}^{(k+1)*\pi}\left|\frac{\sin{t}}{t}\right|dt\geq \int_{k*\pi}^{(k+1)*\pi}\frac{|\sin{t}|}{(k+1)*\pi}dt

Then sum over k=0,1,..,whatevers appropriate. There will be a little left over if x is not a multiple of pi, but this won't matter (you're bounding from below and your integrand is positive).
 

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