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Question Number 132459 by mnjuly1970 last updated on 14/Feb/21

$$....nicecalculus....$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{sin\left(x^2\right)ln\left(x\right)}{x^{\frac{3}{2}}}dx=$$$$solution:$$$$\mathit{\varphi }\stackrel{x^2=t}{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)ln\left(\sqrt{t}\right)}{t^{\frac{3}{4}}}\frac{dt}{t^{\frac{1}{2}}}$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)ln\left(t\right)}{t^{\frac{5}{4}}}dt$$$$\mathit{\psi }\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right).t^a}{t^{\frac{5}{4}}}dt={\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t^{\frac{5}{4}-a}}dt$$$$\mathit{\psi }\left(a\right)=\frac{\mathit{\pi }}{2\mathbf{\Gamma }(\frac{5}{4}-\mathit{a})sin(\frac{5\pi }{8}-\frac{\pi }{2}a)}$$$$\therefore \mathit{\varphi }=\frac{{\mathit{\psi }}^{\prime }\left(0\right)}{4}......$$$${\mathit{\psi }}^{\prime }\left(a\right)=\frac{\pi }{2}\left[\frac{{\mathrm{\Gamma }}^{\prime }(\frac{5}{4}-a)sin(\frac{5\pi }{8}-\frac{\pi a}{2})+\frac{\pi }{2}\mathrm{\Gamma }(\frac{5}{4}-a)cos(\frac{5\pi }{8}-\frac{\mathit{\pi }a}{2})}{{\mathrm{\Gamma }}^2(\frac{5}{4}-a){sin}^2(\frac{5\pi }{8}-\frac{\mathit{\pi }a}{2})}\right]$$$${\mathit{\psi }}^{\prime }\left(0\right)=\frac{\pi }{2}\left(\frac{{\mathrm{\Gamma }}^{\prime }\left(\frac{5}{4}\right)cos\left(\frac{\pi }{8}\right)-\frac{\pi }{2}\mathrm{\Gamma }\left(\frac{5}{4}\right)sin\left(\frac{\pi }{8}\right)}{{\mathrm{\Gamma }}^2\left(\frac{5}{4}\right){cos}^2\left(\frac{\pi }{8}\right)}\right)$$$$=\frac{\pi }{2}\left(\frac{4\psi (1+\frac{1}{4})\mathrm{\Gamma }\left(\frac{1}{4}\right)cos\left(\frac{\pi }{8}\right)-2\pi \mathrm{\Gamma }\left(\frac{1}{4}\right)sin\left(\frac{\pi }{8}\right)}{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right){cos}^2\left(\frac{\pi }{8}\right)}\right)$$$$$$$$\mathit{\varphi }=\frac{\pi }{4}\left(\frac{(2\psi \left(\frac{1}{4}\right)+8)\mathrm{\Gamma }\left(\frac{1}{4}\right)cos\left(\frac{\pi }{8}\right)-\pi \mathrm{\Gamma }\left(\frac{1}{4}\right)sin\left(\frac{\pi }{8}\right)}{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right){cos}^2\left(\frac{\pi }{8}\right)}\right)$$

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