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Question Number 90759 by  M±th+et+s last updated on 25/Apr/20

$$1/{\int }_0^{\mathrm{\infty }}{\int }_x^{\mathrm{\infty }}\frac{cos\left(t\right){ln}^2\left(x\right)}{t\sqrt{x}}dtdx$$$$$$$$2/{\int }_{1/2}^{3/2}ln\left(\mathrm{\Gamma }\left(x\right)\right)dx=\frac{ln\left(\pi \right)-1}{2}$$$$$$$$3/{\int }_0^{\frac{\pi }{2}}cos\left(nt\right){cos}^m\left(t\right)dt=\frac{\pi \mathrm{\Gamma }(m+1)}{2^{m+1}\mathrm{\Gamma }\left(\frac{n+m+2}{2}\right)\mathrm{\Gamma }\left(\frac{2-n+m}{2}\right)}$$$$$$$$4/{\int }_0^{+\mathrm{\infty }}\frac{xexp(\pi cos\left(\pi x\right)sin\left(\pi sin\left(\pi x\right)\right)}{x^2+{\pi }^2}dx=\frac{\pi }{2}exp\left(\pi exp(-{\pi }^2)\right)-1)$$

Commented by abdomathmax last updated on 26/Apr/20

$$1)lettakeatryI={\int }_0^{\mathrm{\infty }}{\int }_x^{\mathrm{\infty }}\frac{cost{ln}^{2}x}{t\sqrt{x}}dtdxbyinversion$$$$wegetI={\int }_0^{\mathrm{\infty }}\left({\int }_0^t\frac{{ln}^{2}x}{\sqrt{x}}\right)\frac{cost}{t}dtbut$$$${\int }_0^t\frac{{ln}^{2}x}{\sqrt{x}}dx{=}_{\sqrt{x}=u}{\int }_0^{\sqrt{t}}\frac{{ln}^2\left(u^2\right)}{u}\left(2u\right)du$$$$=8{\int }_0^{\sqrt{t}}{ln}^{2}uduandbypsrts$$$${\int }_0^{\sqrt{t}}{ln}^{2}udu={\left[u{ln}^{2}u\right]}_0^{\sqrt{t}}-{\int }_0^{\sqrt{t}}u\times \frac{2lnu}{u}du$$$$=\sqrt{t}{ln}^2\left(\sqrt{t}\right)-2{\int }_0^{\sqrt{t}}lnudu$$$$=\frac{1}{4}\sqrt{t}{ln}^2\left(t\right)-2{[ulnu-u]}_0^{\sqrt{t}}$$$$=\frac{1}{4}\sqrt{t}{ln}^2\left(t\right)-2(\sqrt{t}ln\left(\sqrt{t}\right)-\sqrt{t})$$$$=\frac{1}{4}\sqrt{t}{ln}^2\left(t\right)-\sqrt{t}ln\left(t\right)+2\sqrt{t}$$$$=\frac{1}{4}\sqrt{t}{ln}^2\left(t\right)-\sqrt{t}ln\left(t\right)+2\sqrt{t}\Rightarrow$$$${\int }_0^t\frac{{ln}^{2}x}{\sqrt{x}}dx=2\sqrt{t}{ln}^2\left(t\right)-8\sqrt{t}ln\left(t\right)+16\sqrt{t}\Rightarrow$$$$I={\int }_0^{\mathrm{\infty }}(2\sqrt{t}{ln}^2\left(t\right)-8\sqrt{t}ln\left(t\right)+16\sqrt{t})\frac{cost}{t}dt$$$${=}_{\sqrt{t}=z}{\int }_0^{\mathrm{\infty }}(2{zln}^2\left(z^2\right)-8zln\left(z^2\right)+16z)\frac{cos\left(z^2\right)}{z^2}\left(2z\right)dz$$$$={\int }_0^{\mathrm{\infty }}(4{ln}^2\left(z^2\right)-16ln\left(z^2\right)+32)cos\left(z^2\right)dz$$$$=8{\int }_0^{\mathrm{\infty }}{ln}^2\left(z\right)cos\left(z^2\right)dz-32{\int }_0^{\mathrm{\infty }}ln\left(z\right)cos\left(z^2\right)dz$$$$+32{\int }_0^{\mathrm{\infty }}cos\left(z^2\right)dz$$$${\int }_0^{\mathrm{\infty }}cos\left(z^2\right)dz=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-{iz}^2}dz\right)$$$${\int }_0^{\mathrm{\infty }}{e}^{-{iz}^2}dz={\int }_0^{\mathrm{\infty }}{e}^{-{\left(\sqrt{i}z\right)}^2}dz$$$${=}_{\sqrt{i}z=u}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{i}}=\frac{1}{e^{\frac{i\pi }{4}}}\times \frac{\sqrt{\pi }}{2}$$$$=\frac{\sqrt{\pi }}{2}(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})\Rightarrow {\int }_0^{\mathrm{\infty }}cos\left(z^2\right)dz=\frac{\sqrt{\pi }}{2\sqrt{2}}$$$$...becontinuesd...$$

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