Question Number 137419 by mnjuly1970 last updated on 02/Apr/21
$$\:\:\:\:\:\:\:………{mathematical}\:\:\:\:….\:\:\:{analysis}…….. \\ $$$$\:\:\:\:\:\:\:{evaluate}…. \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{\mathrm{2}\pi{x}} −{e}^{\pi{x}} }{{x}\left(\mathrm{1}+{e}^{\mathrm{2}\pi{x}} \right)\left(\mathrm{1}+{e}^{\pi{x}} \right)}{dx}=\lambda\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right){dx}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\lambda\:=\:??? \\ $$$$\:\:\:\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 02/Apr/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{f}\left({ax}\right)−{f}\left({bx}\right)}{{x}}{dx}=\underset{{z}\rightarrow\infty} {\mathrm{lim}}\left({f}\left({z}\right)−{f}\left(\mathrm{0}\right)\right){log}\left(\frac{{a}}{{b}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{\mathrm{2}\pi{x}} −{e}^{\pi{x}} }{{x}\left(\mathrm{1}+{e}^{\mathrm{2}\pi{x}} \right)\left(\mathrm{1}+{e}^{\pi{x}} \right)}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{\mathrm{1}+{e}^{\pi{x}} }−\frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{2}\pi{x}} }}{{x}}{dx}=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right){log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{{log}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\lambda\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\Gamma\left({x}\right)\right){dx}=\lambda\frac{{log}\left(\mathrm{2}\pi\right)}{\mathrm{2}}\:\left({Previous}\:{Examples}\:{on}\:{Community}\right) \\ $$$$\lambda=\frac{{log}\left(\mathrm{2}\right)}{{log}\left(\mathrm{2}\pi\right)}=\frac{\mathrm{1}}{\mathrm{1}+{log}_{\mathrm{2}} \left(\pi\right)} \\ $$
Commented by mnjuly1970 last updated on 02/Apr/21
$$\:{mercey}\:{mr}\:{payan}… \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 02/Apr/21
