$\displaystyle \int_b^c \frac{e^{-ax}*(\sin(bx)-\sin(cx))}{x}\, dx$, $a>0$ and $b,c \in \Bbb R$
I have this improper integral with parameter and I do not understand how to solve it.I have found an answer in my pdf college textbook but the answer is not logical. Here is what it says: $φ(x)= \displaystyle \int_b^c {e^{-ax}*\cos(yx)} \, dy$, $\;x \in (0,+\infty)$
How did this $\cos(yx)$ term appeared from $\sin bx-\sin cx$?
Simply because $$\frac{\mathrm e^{-ax}(\sin cx-\sin bx)}x =\int_b^c {e^{-ax}\cos(yx)} \,\mathrm dy. $$