Change of limits in integral question

[Master J]

Ive come across a few derivations whereby a few changes that don't see so obvious to me occur.
For instance, this one:


Integral, from 0 to infinity, of x.sin(ax), whereby it becomes x.exp(iax) / 2i, from minus infinity to plus. How can the limits be changed like that, and what about the exp(-iax) ?


Cheers chaps!

 

[HallsofIvy]

$$sin(ax)= \frac{e^{iax}- e^{-iax}}{2i}$$
so
$$xsin(ax)= \frac{xe^{iax}}{2i}- \frac{xe^{-iax}}{2i}$$
and then
$$\int_0^\infty xsin(ax)dx= \int_0^\infty \frac{xe^{iax}}{2i}dx- \int_0^\infty\frac{xe^{-iax}}{2i}dx$$

Now, in that second integral, let u= -x. Then du= -dx, when x= 0, u= 0. When $x= \infty$, $u= -\infty$ so that second integral becomes
$$+\int_{-\infty}^0 ue^{iau}du$$.

That is,
$$\int_0^\infty \frac{xe^{iax}}{2i}dx-\int_0^\infty \frac{xe^{-iax}}{2i}dx$$
$$= \int_0^\infty \frac{xe^{iax}}{2i}dx+ \int_{-\infty}^0 \frac{ue^{iau}}{2i}du$$
Since those two integrands are the same, we can combine the two integrals:
[tex]\int_{-\infty}^\infty xe^{iax}dx[/itex]