To compute $\int_{(0,\infty)}e^{(-a+it)x}\frac{\sin x}{x} dx$ In this post there is an intermediate step to compute $  \int_{(0,\infty)}e^{(-c_1+it)x_1}\frac{\sin x_1}{x_1} dx $ where I can't find how to work it out.
To simplify the notation Let's just rename the parameters so the question is to show
$$ I(a):= \int_{(0,\infty)}e^{(-a+it)x}\frac{\sin x}{x} \text{d}x = \frac{1}{2i} \log \left(\frac{a+i(1-t)}{a-i(1+t)}\right)  $$
I've tried so far:
\begin{align*}
I'(a)&:=-\int_{(0,\infty)}e^{(-a+it)x } \sin x  \text{d}x  \\
&= \frac{1}{a-it}e^{(-a+it)x } \sin x  |_0^\infty - \frac{1}{a-it}\int_{(0,\infty)}e^{(-a+it)x } \cos x  \text{d}x \\
&= \frac{1}{(-a+it)^2} e^{(-a+it)x } \cos x |_0^\infty +\frac{1}{(-a+it)^2}\int_{(0,\infty)}e^{(-a+it)x } \sin x  \text{d}x \\
&= -\frac{1}{(-a+it)^2}  -\frac{1}{(-a+it)^2} I'(a)
\end{align*}
So,
$$I'(a) = -\frac{1}{(a-it)^2+1} = \frac{1}{2i} \left(\frac{1}{a-it+i}-\frac{1}{a-it-i}\right)$$
Hence
\begin{align*}
I(a)&=I(0)+\int_0^{a} I'(a) \text{d}a \\
&=I(0)+\frac{1}{2i} \int_0^{a} \left(\frac{1}{a-it+i}-\frac{1}{a-it-i}\right) \text{d}a \\
&=I(0)+\frac{1}{2i} \left.\log \left(\frac{a-it+i}{a-it-i}\right) \right|_0^{a} \\
&=I(0) + \frac{1}{2i} \log \left(\frac{a+i(1-t)}{a-i(1+t)}\right)   -\frac{1}{2i} \log \left(-\frac{1-t}{1+t}\right) 
\end{align*}
Now I'm stuck at $I(0)$.  Note that  $I(0)$ shall be interpreted as
$$\lim\limits_{a\to 0^+}I(a) = \lim\limits_{a\to 0^+} \int_{(0,\infty)}e^{(-a+it)x}\frac{\sin x}{x} \text{d}x $$, however I can't see how to compute this.
And, finally, @metamorphy mentioned that the Frullani integral produces the correct result almost immediately. However I also can't see how.
Pls kindly suggest.
 A: Finding the limit as $a \to 0^+$ of $I(a)$ is, in my opinion, just as hard as finding the general value of $I(a)$. The approach with finding $I'(a)$ is needlessly complex. Instead using that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ you have $$I(a)=\frac{1}{2i}\int_0^{\infty}\frac{e^{(-a+it+i)x}-e^{(-a+it-i)x}}{x}dx$$
With the Frullani integral, if you have $f(x)=e^{-x}$, then the integral is $$I(a)=\frac{1}{2i}\left( f(\infty)-f(0)
 \right)\ln\left( 
\frac{a-it-i}{a-it+i} \right)=\frac{1}{2i}\ln\left( 
\frac{a-it+i}{a-it-i} \right)$$
A: Let $s=-a+it$.
We now have
$$F(s) = \int_0^\infty e^{-sx} \frac{\sin x}{x}\, dx .$$
Differentiating
$$F^\prime(s) =  -\int_0^\infty e^{-sx} \sin x\, dx. $$
This is a familiar Laplace transform!
$$F^\prime(s) = - \frac{1}{1+s^2}.$$
$$F(s) = \text{const.} - \arctan(s).$$
Since it is well-known that $F(0) = \frac{\pi}{2}$, we have
$$F(s) = \frac{\pi}{2} - \arctan(s).$$
$$F(-a+it) = \frac{\pi}{2} - \arctan(-a+it).$$
You can write this in terms of logarithms, if you like:
$$-a+it = -\tan \left(w-\frac{\pi}{2}\right)= \cot w = - i \frac{e^{iw}+e^{-iw}}{e^{iw}-e^{-iw}}.$$
Simplifying and solving the resulting quadratic equation in $e^{iw}$ yields the expression in terms of the logarithm function.
$$F(-a+it) = \frac{1}{2i} \log \left[ \frac{a-(t-1)i}{a-(t+1)i} \right].$$
A: After obtaining $I'(a) = -\frac{1}{(a-it)^2+1}$, proceed as follows
\begin{align*}
I(a)&=I(\infty)+\int_\infty^{a} I'(s) ds 
=0- \int_\infty^{a} \frac{ds}{(s-it)^2+1}= \frac{1}{2i} \log \frac{a+i(1-t)}{a-i(1+t)}
\end{align*}
