Plz don't mislead .

roopa 1991 has given a completely wrong condition on integrability .It has nothing to do with continuity of the fn at a single point .

All of u have integrated ( defnite integral ) box fn . Now that's  discontinuous at every integral point ,yet the definite integral exists.

The qn is whether we can integrate it by using a closed form formula or not ( note Si(x) is not a closed form fn . )

And in this regards ,I can say that this indefinite integral can't be integrated in a closed form formula using elementary functions .

Yet definite integral of this fn in the range ( -inf to inf ) or (0 to inf ) has a closed form formula ( pi and pi/2 respectively ).