I am currently learning contour integration and I noticed that certain integrals are not mentioned at all when discussing contour integration and some of those are the following : $$\int_{0}^{\infty} {e^{-x}\ln(x)}\,dx = -\gamma$$ $$\int_{0}^{\infty} {\frac{\sin(x)\ln(x)}{x}}\,dx = -\frac{\gamma\pi}{2}$$ $$\mathcal L^{-1} \left\{\frac{\log(s)}{s}\right\}(t) =$$ $$\frac{1}{2\pi i}\int_{\sigma-\infty i}^{\sigma+\infty i} {\left({\log(s) \over s}\right)e^{st}}\,ds =$$ $$-\gamma-\ln(t)$$ $$\int_{0}^{1} {\frac{\ln(x)}{1+x^{2}}}\,dx = -\mathit C$$ All of them when anyone I saw try to solve them they don't use contour integration so I tried solving them myself but I don't think I have enough knowledge to solve them, so is it possible to solve them using contour integration ? And to note I already know how to solve them through other means but I am Interested in solving them using contour integration.
I will show one of my attempts to solve the first one, I initially forgot that keyhole contours are mainly used when dealing with polynomials raised to a fraction so I defined my complex function as $$f(z) = \frac{\log(z)}{e^z}$$ and chose this contour
Which is the usual keyhole contour : $\epsilon\to R\to (0,2\pi)\to R\to \epsilon$
As $\epsilon\to 0$ and $R\to \infty$
So we have a total of four paths creating our closed contour, and since there's no poles or branch cuts inside the contour then its value is 0, and according to Jordan's lemma the big circle will tend to 0 and I parameterized the smaller circle and found its value to be also 0 so then what left is just the two paths going from infinity to 0 and vice versa so what we will be left with is just the difference of the argument of z since I choose the branch cut to be $(0,2\pi)$ so then we have $0 = -2\pi i$
So I just messed up, I didn't recover the original integral nor did I do the calculation right so I would appreciate it if someone would tell me what I did wrong up there and propose another way if possible.
As mentioned in the comments that Jordan's lemma doesn't hold yes I am now convinced but the big circle will nonetheless still converge into zero if we just parameterized it like this $z = Re^\left(i\theta\right)$.