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Question Number 144432 by ArielVyny last updated on 25/Jun/21

$${\int }_0^{\propto }{t}^{n-2}costdt$$

Answered by Dwaipayan Shikari last updated on 25/Jun/21

$${\int }_0^{\mathrm{\infty }}{t}^{s-1}cos\left(t\right)dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{it}+t^{s-1}{e}^{-it}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}(i^s{u}^{s-1}{e}^{-u}du+{(-i)}^s{j}^{s-1}{e}^{-j}dj)$$$$=\frac{\mathrm{\Gamma }\left(s\right)}{2}(i^s+{(-i)}^s)=\mathrm{\Gamma }\left(s\right)cos\left(\frac{\pi s}{2}\right)$$$$s=n-1{\int }_0^{\mathrm{\infty }}{t}^{n-2}cos\left(t\right)dt=\mathrm{\Gamma }(n-1)cos\left(\frac{\pi (n-1)}{2}\right)$$

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