I just came across these four integrals, evaluated them, and noticed they show this weird alternate pattern, I think its cool but I have no idea why they do. Anyone knows an intuitive reason for this?
$$\hspace{1cm} \int_0^\infty \frac{\cos(\ln(x))}{1+x}\mathrm{d}x=0 \hspace{2cm} \int_0^\infty \frac{\sin(\ln(x))}{1+x}\mathrm{d}x= \frac{\pi}{\sinh(\pi)}$$
$$\int_0^\infty \frac{\cos(\ln(x))}{(1+x)^2}\mathrm{d}x= \frac{\pi}{\sinh(\pi)} \hspace{1cm} \int_0^\infty \frac{\sin(\ln(x))}{(1+x)^2}\mathrm{d}x=0 $$
Best Answer
Consider the integral
$$ I = \int_{0}^{\infty} \frac{e^{i\ln(x)}}{1+x} dx= \int_{0}^{\infty} \frac{x^i}{1+x}dx $$
Do the substitution $\displaystyle w = \frac{1}{1+x}$
$$I = \int_{0}^{1} (1-w)^{i}w^{-1-i} dw = B(1+i,-i) = \Gamma(1+i)\Gamma(-i) = -\frac{\pi}{\sin(\pi i)} = \frac{i\pi}{\sinh(\pi)}$$
Note that $\displaystyle \frac{\pi}{\sinh(\pi)}$ is real valued. Then take the real and imaginary parts:
$$ \Im \int_{0}^{\infty} \frac{e^{i\ln(x)}}{1+x} dx= \int_{0}^{\infty} \frac{\cos(\ln(x))}{1+x} dx = 0 $$
$$ \Re \int_{0}^{\infty} \frac{e^{i\ln(x)}}{1+x} dx= \int_{0}^{\infty} \frac{\sin(\ln(x))}{1+x} dx = \frac{\pi}{\sinh(\pi)} $$
Edit: Similarly, if we make the same substitution $\displaystyle w = \frac{1}{(1+x)}$
\begin{align*}\int_{0}^{\infty} \frac{e^{i\ln(x)}}{(1+x)^2}dx = &\int_{0}^{\infty} \frac{x^i}{(1+x)^2}dx=\int_{0}^{1} (1-w)^{i}w^{-i} dw \\ =& \Gamma\left(1+i\right)\Gamma\left(1-i\right) \\ =& i\Gamma(i)\Gamma(1-i)\\ =& \frac{i\pi}{\sin\left(i\pi\right)}\\ =& \frac{\pi}{\operatorname{sinh}\left(\pi\right)}\end{align*}
This last is a real value
Take real and imaginary parts.
So
$$ \Re \int_{0}^{\infty} \frac{e^{i\ln(x)}}{(1+x)^2}dx = \int_{0}^{\infty} \frac{\cos(\ln(x))}{(1+x)^2}dx = \frac{\pi}{\operatorname{sinh}\left(\pi\right)}$$
$$ \Im \int_{0}^{\infty} \frac{e^{i\ln(x)}}{(1+x)^2}dx = \int_{0}^{\infty} \frac{\sin(\ln(x))}{(1+x)^2}dx = 0$$
Edit 2: Another approach for the second case.
\begin{align*} \int_{0}^{1} \frac{\sin(\ln(x))}{(1+x)^2} dx =& \frac{1}{2i} \left[\int_{0}^{\infty} \frac{x^i}{(1+x)^2} dx - \int_{0}^{\infty} \frac{x^{-i}}{(1+x)^2} dx\right]\\ =& \frac{1}{2i}\left[\int_{0}^{\infty} \frac{x^i}{(1+x)^2} dx - \underbrace{\int_{0}^{\infty} \frac{w^i}{(1+w)^2}dw}_{w \mapsto \frac{1}{x}}\right]\\ =& 0 \end{align*}
\begin{align*} \int_{0}^{1} \frac{\cos(\ln(x))}{(1+x)^2} dx =& \frac{1}{2} \left[\int_{0}^{\infty} \frac{x^i}{(1+x)^2} dx + \int_{0}^{\infty} \frac{x^{-i}}{(1+x)^2} dx\right]\\ =& \frac{1}{2}\left[\int_{0}^{\infty} \frac{x^i}{(1+x)^2} dx + \underbrace{\int_{0}^{\infty} \frac{w^i}{(1+w)^2}dw}_{w \mapsto \frac{1}{x}}\right]\\ =& \int_{0}^{\infty} \frac{x^i}{(1+x)^2} dx\\ =& \frac{\pi}{\sinh(\pi)} \end{align*}