User:Bitcoin123

$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-xy}\cdot\frac{\ln\left(xy^{1/\phi}\right)}{\ln(xy)}\mathrm dxdy=1-\frac{\gamma}{\phi^2}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-xy}\cdot\frac{\ln\left(xy^{\phi}\right)}{\ln(xy)}\mathrm dxdy=1+\frac{\gamma}{\phi}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x}{1-xy}\cdot\frac{\ln\left(xy^{\phi^2}\right)}{\ln(xy)}\mathrm dxdy=1+\phi\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x\ln(x^ay^{a+1})}{(1-xy)\ln(xy)}\mathrm dxdy=a+\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{xy(x+y)\ln(xy^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=\phi$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(1+xy)(x+y)\ln(xy^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=\phi^4-2$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(1+2xy)(x+y)\ln(xy^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=4\phi$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(c+dxy)(x+y)\ln(xy^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=(2c+d)\phi$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(c+dxy)(x+y)\ln(x/y^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=-\frac{(2c+d)}{\phi}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(1-3xy)(x+y)\ln(xy^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=-\phi$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(1-3xy)(x+y)\ln(x/y^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=\frac{1}{\phi}$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(1+\sqrt{5}xy)(x+y)\ln(xy^{\sqrt{5}})}{(1-xy)\ln(xy)}\mathrm dxdy=\phi^4$$

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$$\int_{0}^{1}\ln(x)\ln\left({\frac{x^{\phi}(-\ln(x))^{\phi^3}}{[1-\ln(x)]^{-\gamma\phi^3}}}\right)\mathrm dx=\phi-\phi^2$$

$$\int_{0}^{1}\ln(x)\ln\left({\frac{x^{(n-1)/2}(-\ln(x))^{n}}{1-\ln(x)}}\right)\mathrm dx=n\gamma$$

$$\int_{0}^{1}\ln(x)\ln\left({\frac{x^{(n-1)/2}(-\ln(x))^{n}}{[1-\ln(x)]^m}}\right)\mathrm dx=n\gamma+m-1$$

$$\int_{0}^{1}\ln(x)\ln\left({\frac{x^{(n-1)/2}(-\ln(x))^{n}}{[1-\ln(x)]^{-n\gamma}}}\right)\mathrm dx=-1$$

$$\sum_{n=2}^{\infty}(n-1)(n-2)\cdots(n-k)(\zeta(n)-1)=k!\zeta(k+1)$$

$$\sum_{n=2}^{\infty}\frac{(n-1)^3}{n}(\zeta(n)-1)=2\zeta(3)+\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x\ln y}{(1-xy)\ln(xy)}\mathrm dxdy=\gamma$$

$$\int_{0}^{1}\int_{0}^{1}\frac{x\ln y}{(1+xy)\ln(xy)}\mathrm dxdy=\ln\left(\frac{\pi}{4 }\right)$$

$$\int_{0}^{1}\int_{0}^{1}\frac{(x+y)\ln(x^ay^b)}{(1-xy)\ln(xy)}\mathrm dxdy=a+b$$

$$\int_{0}^{1}\int_{0}^{1}\frac{xy(x+y)\ln(x^ay^b)}{(1-xy)\ln(xy)}\mathrm dxdy=\frac{a+b}{2}$$