20 yo Math's artist. Some of my creations :

$\begin{align} \mathrm{If}\;\;\;\psi : s \mapsto \frac{\Gamma'(s)}{\Gamma(s)} \;,\;a>0\;\mathrm{and}\;b>0 \;\;\;,\; \mathrm{Then}: \newline \newline \int_{0}^{\infty} \frac{\psi(1+az)}{\cosh(b\pi z)}\,dz-\int_{0}^{\infty} \frac{\psi(1+bz)}{\cosh(a\pi z)}\,dz&=\frac{1}{2}\ln\left(\sqrt[a]{\frac{a \, \omega^2}{b}}\sqrt[b]{\frac{a }{b\, \omega^2}}\right) \;\;\;,\;\;\;\; \omega=\frac{\Gamma\left(\frac{1}{4}\right)}{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}\\ \int_{0}^{\infty} \frac{\psi\left(1+\frac{\sqrt{3}}{2}az\right)}{2\cosh(b\pi z)+1}\,dz-\int_{0}^{\infty} \frac{\psi\left(1+\frac{\sqrt{3}}{2}bz\right)}{2\cosh(a\pi z)+1}\,dz&=\frac{1}{3\sqrt{3}}\ln\left(\sqrt[a]{\frac{a \, \tau^3}{b}}\sqrt[b]{\frac{a }{b\, \tau^3}}\right) \;\;\;,\;\;\;\; \tau=\frac{\Gamma\left(\frac{1}{3}\right)}{\sqrt[6]{3}\Gamma\left(\frac{2}{3}\right)}\\ \newline \newline \newline \end{align}$ $\begin{align} \int_{-\infty}^{\infty} \frac{\ln\left(\frac{\tanh(z)}{z}\right)}{z^2(1+z^2)}\,dz= 2\pi \ln\left(\frac{\mathrm{B}(\frac{1}{\pi},\frac{1}{\pi})}{2\pi}\right)\;\;\;,\;\;\;\; \mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ \end{align}$

$\begin{align} \int_{-\infty}^{\infty} \frac{\sinh(\pi z)}{z(1+z^2)(\cosh(\pi z)+1)^2}\,dz= \frac{\zeta(3)}{\zeta(2)}\\ \end{align}$

$\begin{align} \int_{-\infty}^{\infty} \frac{\sinh(\pi z)}{z(1+z^2)(\cosh(\pi z)+1)^3}\,dz=\frac{1}{6}\left( \frac{\zeta(3)}{\zeta(2)}+\frac{\zeta(5)}{\zeta(4)}\right) \end{align}$

$\begin{align} \prod_{n=1}^{\infty} \left(\frac{(2n+1)^{(2n+1)\ln\left(2n+1\right)}}{(2n)^{n\ln\left(2n\right)}(2n+2)^{(n+1)\ln(2n+2)}}\right)^{(-1)^n}=\sqrt[\pi]{\pi^{4 G}}\, \sqrt[\pi]{\frac{e^{14 \gamma G}}{16^G e^{4 \beta'(2)}e^{4G}}} 2^{\ln\left(2\right)} \end{align}$

$\begin{align} \prod_{n=1}^{\infty} \left(\frac{2(n+1)^2}{e^3}\left(\frac{4n(n+1)}{e^3}\right)^{2n\ln\left(\left(\frac{n}{1+n}\right)^{n+1}e\right)}\right)^{(-1)^{n+1}}=\frac{4\sqrt[\pi^2]{\pi^{14 \zeta(3)}}}{e^3}\, \sqrt[\pi^2]{\frac{e^{14 \gamma \zeta(3)}}{2^{16\zeta(3)} e^{14 \zeta'(3)}}} \end{align}$

$\begin{align} \int_{0}^{\infty} \frac{\ln(\cosh(cz)+\cos(\theta))}{(a^2+z^2)(b^2+z^2)}\,dz =\frac{\pi ^2}{a^2-b^2} \ln\left(\frac{\sqrt[a]{\Gamma\left(\frac{1}{2}+\frac{ac+\theta}{2\pi}\right)\Gamma\left(\frac{1}{2}+\frac{ac-\theta}{2\pi}\right)}}{\sqrt[b]{\Gamma\left(\frac{1}{2}+\frac{bc+\theta}{2\pi}\right)\Gamma\left(\frac{1}{2}+\frac{bc-\theta}{2\pi}\right)}} \frac{\sqrt[b]{\sqrt{2\pi}}}{\sqrt[a]{\sqrt{2\pi}}}\right) \end{align}$

$\begin{align} \int_{0}^{\infty} \frac{\tan^{-1}\left(\tanh\left(\frac{bz}{2}\right)\tan\left(\frac{\theta}{2}\right)\right)}{z(a^2+z^2)}\,dz =\frac{\pi}{2a^2} \ln\left(\frac{\Gamma\left(\frac{1}{2}+\frac{ab+\theta}{2\pi}\right)}{\Gamma\left(\frac{1}{2}+\frac{ab-\theta}{2\pi}\right)} \frac{\Gamma\left(\frac{1}{2}-\frac{\theta}{2\pi}\right)}{\Gamma\left(\frac{1}{2}+\frac{\theta}{2\pi}\right)}\right) \end{align}$

Instagram @elies.calculus :)