Abs Value of Sine Integral

[sparkster]

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.

 

[dextercioby]

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.

 

[sparkster]

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.

 

[dextercioby]

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

Daniel.

 

[shmoe]

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).