Weird $\sin(\ln (x))$, $\cos(\ln (x))$ pattern

Question

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?

$$\hspace{1cm} \int_0^\infty \frac{\cos(\ln(x))}{1+x}\mathrm{d}x=0 \hspace{2cm} \int_0^\infty \frac{\sin(\ln(x))}{1+x}\mathrm{d}x= \frac{\pi}{\sinh(\pi)}$$ $$\int_0^\infty \frac{\cos(\ln(x))}{(1+x)^2}\mathrm{d}x= \frac{\pi}{\sinh(\pi)} \hspace{1cm} \int_0^\infty \frac{\sin(\ln(x))}{(1+x)^2}\mathrm{d}x=0 $$

Answer 1

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*}

Answer 2

For at least the fourth integral quoted in the question, the zero value is proved by elementary means.

Define $u=1/x$, then

$\int_0^\infty\dfrac{\sin[\ln(x)]dx}{(1+x)^2}=\int_\infty^0\dfrac{\sin[-\ln(u)][-du/u^2]}{[1+(1/u)]^2}=\int_\infty^0\dfrac{\sin[\ln(u)]du}{(1+u)^2}.$

Noting the switched limits, the integral on the right is the negative of the equal integral on the left, thus zero if it converges. But the integrand (despite its ill behavior) is bounded as $x\to0^+$, and comparable with $1/x^2$ as $x\to\infty$; thus both ends converge (absolutely).

Answer 3

The first two integrals don't converge.

The easiest way to see that they don't converge is to make the substitution $u = \ln (x)$.

The first integral then becomes $$\int_{-\infty}^{\infty} \frac{\cos (u)}{1+e^{u}} \, e^{u} \, \mathrm du, $$ and the second integral becomes $$\int_{-\infty}^{\infty} \frac{\sin (u)}{1+e^{u}} \, e^{u} \, \mathrm du. $$

These are not convergent integrals since the integrand of the first integral oscillates like $\cos(u)$ as $u \to + \infty$, while the integrand of the second integral oscillates like $\sin(u)$ as $u \to + \infty$.

Wolfram Alpha, however, claims that $$\int_{0}^{\infty} \frac{x^{i}}{1+x} \,\mathrm dx = \int_{0}^{\infty} \frac{x^{(1+i)-1}}{(1+x)^{(1+i)-i }} \, \mathrm dx = i \pi \operatorname{csch}(\pi). $$

I assume it's using the integral formula $$\int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}} \, \mathrm dx = B(a, b). $$

But this formula is only valid if $\Re(a)>0$ and $\Re(b) >0$.