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Question Number 207582 by sniper237 last updated on 19/May/24

$${\int }_0^{\pi }ln\left(sinx\right)dx=-\pi ln2$$$${\int }_0^{1}ln\mathrm{\Gamma }\left(x\right)dx=ln\left(2\pi \right)$$

Answered by mathzup last updated on 21/May/24

$$I={\int }_0^{\pi }ln\left(sinx\right)dx={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx$$$$+{\int }_{\frac{\pi }{2}}^{\pi }ln\left(sinx\right)dx(\to x=\frac{\pi }{2}+t)$$$$=-\frac{\pi }{2}ln\left(2\right)+{\int }_0^{\frac{\pi }{2}}ln\left(cost\right)dt$$$$=-\frac{\pi }{2}ln2-\frac{\pi }{2}ln2=-\pi ln2$$

Answered by mathzup last updated on 21/May/24

$$\mathrm{\Gamma }\left(x\right).\mathrm{\Gamma }(1-x)=\frac{\pi }{sin\left(\pi x\right)}\Rightarrow$$$$ln\left(\mathrm{\Gamma }\left(x\right)\right)+ln\left(\mathrm{\Gamma }(1-x)\right)=ln\pi -ln\left(sin\left(\pi x\right)\right)$$$$\Rightarrow {\int }_0^{1}ln\left(\mathrm{\Gamma }\left(x\right)\right)dx+{\int }_0^{1}ln\left(\mathrm{\Gamma }(1-x)\right)dx$$$$=ln\left(\pi \right)-{\int }_0^{1}ln\left(sin\left(\pi x\right)\right)dx(\to \pi x=t)$$$$=ln\left(\pi \right)-{\int }_0^{\pi }ln\left(sint\right)\frac{dt}{\pi }$$$$=ln\left(\pi \right)-\frac{1}{\pi }(-\pi ln2)$$$$=ln\left(\pi \right)+ln\left(2\right)=ln\left(2\pi \right)$$$$but{\int }_0^{1}ln\left(\mathrm{\Gamma }(1-x)\right)dx{=}_{1-x=t}{\int }_1^{0}ln\left(\mathrm{\Gamma }\left(t\right)\right)(-dt)$$$$={\int }_0^{1}ln(\mathrm{\Gamma }\left(x\right)dx\Rightarrow$$$${\int }_0^{1}ln\left(\mathrm{\Gamma }\left(x\right)\right)dx=\frac{1}{2}ln\left(2\pi \right)$$$$ilyaerreurdanslenonce!...$$

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