cost(xt)/1+t dt anyone can help
Lets start evaluating the indefinite integral...
$\displaystyle \varphi (x, t)= \int \frac{e^{x t}}{1+t}\cdot dt$ (1)
Setting $\displaystyle 1+t=\tau$ You obtain first the integral...
$\displaystyle \varphi (x,\tau) = e^{-x} \int \frac{e^{x \tau}}{\tau}\cdot d\tau$ (2)
... and now setting $\displaystyle x \tau = \xi$...
$\displaystyle \varphi (x,\xi)= e^{-x} \int \frac{e^{\xi}}{\xi}\cdot d\xi = e^{-x}\cdot Ei(\xi) + c$ (3)
... where $\displaystyle Ei(*)$ is the socalled 'Integral exponential function'. What we have obtained is that...
$\displaystyle \int \frac{e^{x t}}{1+t}\cdot dt = e^{-x}\cdot Ei [x\cdot (1+t)] + c$ (4)
If now we write in (4) $\displaystyle i x$ instead of $\displaystyle x$ we obtain...
$\displaystyle \int \frac{\cos x t}{1+t}\cdot dt = Re \{ e^{-i x}\cdot Ei[i x \cdot (1+t)]\} + c$ (5)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$