integration by parts on a fourier transformation

Question

Given is a function $$ \dfrac{1}{2 \pi} \int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f(t)}{g(-t/h)} \, dt.$$ In the paper I'm working with it's written that by integration by parts, we have $$ \dfrac{1}{2 \pi} \int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f(t)}{g(-t/h)} \, dt \\ = \dfrac{1}{2 \pi i x}\int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f'(t)}{g(-t/h)} \, dt + \dfrac{1}{2 \pi i x h} \int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f(t)g'(-t/h)}{(g(-t/h))^2} \, dt.$$ I know that the formula for integration by parts is $\int uv' = uv - \int u'v$. So my question is: "What is $u$ and what is $v$ in my example?" Firstly, if you compare the formula with the calculation, you will see that in the formula we have only one integral but in the calculation two integrals, which confuses me. Secondly I don't know why we have the "$ix$" in the denominator in the two parts. It's obvious, that in the first part $f$ was derived and in the second part $g$. But why is $\mathrm{e}^{-itx}$ derived in the two parts? Thanks!

Answer 1

If you pick $u(t):=\mathrm{e}^{-itx} / (-ix)$ and $v(t) := f(t) / g(-t/h)$, then $u'(t):=\mathrm{e}^{-itx}$ and \begin{equation} v'(t) = \frac{f'(t)}{g(-t/h)}+f(t)\cdot\bigg(- \frac{1}{\big(g(-t/h)\big)^2}\bigg) \cdot g'(-t/h) \cdot (-1/h). \end{equation} If follows that \begin{align*} \dfrac{1}{2 \pi} \int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f(t)}{g(-t/h)} \, \mathrm{d}t &= \dfrac{1}{2 \pi} \int\limits_{- \infty}^{+ \infty} u'(t)v(t) \, \mathrm{d}t \\ &= - \frac{1}{2\pi} \int\limits_{- \infty}^{+ \infty} u(t) v'(t) \mathrm{d}t \\ &=\dfrac{1}{2 \pi i x}\int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f'(t)}{g(-t/h)} \, \mathrm{d}t + \dfrac{1}{2 \pi i x h} \int\limits_{- \infty}^{+ \infty} \mathrm{e}^{-itx}\dfrac{f(t)g'(-t/h)}{(g(-t/h))^2} \, \mathrm{d}t. \end{align*}