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Question Number 138041 by mnjuly1970 last updated on 09/Apr/21

$$.......advanced.........calculus......$$$$provethat:::$$$$\mathit{\varphi }={\int }_0^1\psi \left(x\right).sin\left(2\pi x\right)dx=-\frac{\pi }{2}...✓$$$$$$

Answered by Dwaipayan Shikari last updated on 09/Apr/21

$${\int }_0^1\psi \left(x\right)sin\left(2\pi x\right)dx$$$$={\left[sin\left(2\pi x\right)log\left(\mathrm{\Gamma }\left(x\right)\right)\right]}_0^1-2\pi {\int }_0^{1}log\left(\mathrm{\Gamma }\left(x\right)\right)cos\left(2\pi x\right)$$$$=-\pi {\int }_0^{1}log\left(\pi \right)cos\left(2\pi x\right)+\pi {\int }_0^{1}log\left(sin\left(\pi x\right)\right)cos\left(2\pi x\right)dx$$$$={\int }_0^{\pi }log\left(sin\left(u\right)\right)cos\left(2u\right)du$$$$={\left[\frac{1}{2}log\left(sin\left(u\right)\right)sin\left(2u\right)\right]}_0^{\pi }-\frac{1}{2}{\int }_0^{\pi }\frac{sin\left(2u\right)cos\left(u\right)}{sin\left(u\right)}du$$$$=-\frac{1}{2}{\int }_0^{\pi }1+cos\left(2u\right)du=-\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 09/Apr/21

$$verynicemr$$$$payan..thankyousomuch..$$

Answered by mnjuly1970 last updated on 09/Apr/21

$$\mathit{\varphi }={\int }_0^1\psi \left(x\right).sin\left(2\pi x\right)dx$$$$\mathit{\varphi }=-{\int }_0^1\psi (1-x).sin\left(2\pi x\right)dx$$$$2\mathit{\varphi }={\int }_0^1\{(\psi \left(x\right)-\psi (1-x)\}sin\left(2\pi x\right)dx$$$$=-\pi {\int }_0^{1}cot\left(\pi x\right).sin\left(2\pi x\right)dx$$$$=-2\pi {\int }_0^1\frac{cos\left(\pi x\right)}{sin\left(\pi x\right)}\{sin\left(\pi x\right).cos\left(\pi x\right)\}dx$$$$\therefore \mathit{\varphi }=-\pi {\int }_0^1\frac{1+cos\left(2\pi x\right)}{2}dx$$$$=-\frac{\pi }{2}.....✓$$

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