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Question Number 137419 by mnjuly1970 last updated on 02/Apr/21

$$.........mathematical....analysis........$$$$evaluate....$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{e^{2\pi x}-e^{\pi x}}{x(1+e^{2\pi x})(1+e^{\pi x})}dx=\lambda {\int }_0^{1}ln(\mathrm{\Gamma }\left(x\right)dx$$$$\lambda =???$$$$$$

Answered by Dwaipayan Shikari last updated on 02/Apr/21

$${\int }_0^{\mathrm{\infty }}\frac{f\left(ax\right)-f\left(bx\right)}{x}dx=\underset{z\to \mathrm{\infty }}{\lim }(f\left(z\right)-f\left(0\right))log\left(\frac{a}{b}\right)$$$${\int }_0^{\mathrm{\infty }}\frac{e^{2\pi x}-e^{\pi x}}{x(1+e^{2\pi x})(1+e^{\pi x})}dx={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{1+e^{\pi x}}-\frac{1}{1+e^{2\pi x}}}{x}dx=(-\frac{1}{2})log\left(\frac{1}{2}\right)$$$$=\frac{log\left(2\right)}{2}$$$$\lambda {\int }_0^{1}log\left(\mathrm{\Gamma }\left(x\right)\right)dx=\lambda \frac{log\left(2\pi \right)}{2}\left(PreviousExamplesonCommunity\right)$$$$\lambda =\frac{log\left(2\right)}{log\left(2\pi \right)}=\frac{1}{1+{log}_2\left(\pi \right)}$$

Commented by mnjuly1970 last updated on 02/Apr/21

$$merceymrpayan...$$$$$$

Commented by mnjuly1970 last updated on 02/Apr/21

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