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Question Number 214750 by universe last updated on 18/Dec/24

Answered by aleks041103 last updated on 21/Dec/24

$$J={\int }_0^{\mathrm{\infty }}{\int }_z^{\mathrm{\infty }}{\int }_z^{\mathrm{\infty }}cos(x^2-y^2)dxdydz$$$$cos(x^2-y^2)=cos\left(x^2\right)cos\left(y^2\right)+sin\left(x^2\right)sin\left(y^2\right)$$$$\Rightarrow {\int }_z^{\mathrm{\infty }}{\int }_z^{\mathrm{\infty }}cos(x^2-y^2)dxdy=$$$$={\int }_z^{\mathrm{\infty }}{\int }_z^{\mathrm{\infty }}[cos\left(x^2\right)cos\left(y^2\right)+sin\left(x^2\right)sin\left(y^2\right)]dxdy=$$$$=A^2+B^2$$$$where$$$$A={\int }_z^{\mathrm{\infty }}cos\left(x^2\right)dx=\sqrt{\frac{\pi }{2}}{\int }_z^{\mathrm{\infty }}cos\left(\frac{\pi }{2}{\left(\sqrt{\frac{2}{\pi }}x\right)}^2\right)d\left(\sqrt{\frac{2}{\pi }}x\right)$$$$B={\int }_z^{\mathrm{\infty }}sin\left(x^2\right)dx=\sqrt{\frac{\pi }{2}}{\int }_z^{\mathrm{\infty }}sin\left(\frac{\pi }{2}{\left(\sqrt{\frac{2}{\pi }}x\right)}^2\right)d\left(\sqrt{\frac{2}{\pi }}x\right)$$$$weknow$$$$C\left(t\right):={\int }_0^{t}cos\left(\frac{\pi x^2}{2}\right)dx$$$$S\left(t\right):={\int }_0^{t}cos\left(\frac{\pi x^2}{2}\right)dx$$$$whereSandCareFresnelintegrals$$$$\Rightarrow A=\sqrt{\frac{\pi }{2}}[C\left(\mathrm{\infty }\right)-C\left(\sqrt{\frac{2}{\pi }}z\right)]$$$$\Rightarrow B=\sqrt{\frac{\pi }{2}}[S\left(\mathrm{\infty }\right)-S\left(\sqrt{\frac{2}{\pi }}z\right)]$$$$itiswellknownthat$$$$C\left(\mathrm{\infty }\right)=S\left(\mathrm{\infty }\right)=\frac{1}{2}$$$$\Rightarrow A=\sqrt{\frac{\pi }{2}}[\frac{1}{2}-C\left(\sqrt{\frac{2}{\pi }}z\right)]$$$$\Rightarrow B=\sqrt{\frac{\pi }{2}}[\frac{1}{2}-S\left(\sqrt{\frac{2}{\pi }}z\right)]$$$$$$$$\Rightarrow A^2+B^2=\frac{\pi }{2}[\frac{1}{2}+S^2\left(\sqrt{\frac{2}{\pi }}z\right)+C^2\left(\sqrt{\frac{2}{\pi }}z\right)-S\left(\sqrt{\frac{2}{\pi }}z\right)-C\left(\sqrt{\frac{2}{\pi }}z\right)]=f\left(\sqrt{\frac{2}{\pi }}z\right)$$$$\Rightarrow J={\int }_0^{\mathrm{\infty }}f\left(\sqrt{\frac{2}{\pi }}z\right)dz=$$$$=\sqrt{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}f\left(z\right)dz$$$$\Rightarrow J={\left(\frac{\pi }{2}\right)}^{3/2}{\int }_0^{\mathrm{\infty }}[{(\frac{1}{2}-C\left(z\right))}^2+{(\frac{1}{2}-S\left(z\right))}^2]dz$$$$laterwewillshow:$$$${\int }_0^{\mathrm{\infty }}{(\frac{1}{2}-C\left(z\right))}^{2}dz=\frac{1}{2\sqrt{2}\pi }$$$${\int }_0^{\mathrm{\infty }}{(\frac{1}{2}-S\left(z\right))}^{2}dz=\frac{4-\sqrt{2}}{4\pi }$$$$\Rightarrow J=(1-\frac{3\sqrt{2}}{4})\sqrt{\frac{\pi }{8}}$$

Commented by aleks041103 last updated on 21/Dec/24

$$\int {(\frac{1}{2}-C\left(z\right))}^{2}dz=$$$$=z{(\frac{1}{2}-C\left(z\right))}^2+2\int z(\frac{1}{2}-C\left(z\right))C{}^{\prime }\left(z\right)dz$$$$\int z(\frac{1}{2}-C\left(z\right))C{}^{\prime }\left(z\right)=$$$$=\int zcos\left(\frac{\pi }{2}{z}^2\right)(\frac{1}{2}-C\left(z\right))dz=$$$$=\frac{1}{\pi }\int (\frac{1}{2}-C\left(z\right))d\left(sin\left(\frac{\pi z^2}{2}\right)\right)=$$$$=\frac{1}{\pi }sin\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-C\left(z\right))+\frac{1}{\pi }\int sin\left(\frac{\pi z^2}{2}\right)C{}^{\prime }\left(z\right)dz$$$$\int sin\left(\frac{\pi z^2}{2}\right)C{}^{\prime }\left(z\right)dz=\int sin\left(\frac{\pi z^2}{2}\right)cos\left(\frac{\pi z^2}{2}\right)dz=$$$$=\frac{1}{2\sqrt{2}}\int sin\left(\frac{\pi {\left(\sqrt{2}z\right)}^2}{2}\right)d\left(\sqrt{2}z\right)=\frac{1}{2\sqrt{2}}S\left(\sqrt{2}z\right)$$$$\Rightarrow \int z(\frac{1}{2}-C\left(z\right))C{}^{\prime }\left(z\right)=$$$$=\frac{1}{\pi }sin\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-C\left(z\right))+\frac{S\left(\sqrt{2}z\right)}{2\sqrt{2}\pi }$$$$\Rightarrow F\left(z\right)=\int {(\frac{1}{2}-C\left(z\right))}^{2}dz=$$$$=z{(\frac{1}{2}-C\left(z\right))}^2+\frac{2}{\pi }sin\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-C\left(z\right))+\frac{S\left(\sqrt{2}z\right)}{\sqrt{2}\pi }$$$$F\left(0\right)=0$$$$F\left(\mathrm{\infty }\right)=\frac{S\left(\mathrm{\infty }\right)}{\sqrt{2}\pi }=\frac{1}{2\sqrt{2}\pi }$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{(\frac{1}{2}-C\left(z\right))}^{2}dz=\frac{1}{2\sqrt{2}\pi }$$

Commented by aleks041103 last updated on 21/Dec/24

$$\int {(\frac{1}{2}-S\left(z\right))}^{2}dz=$$$$=z{(\frac{1}{2}-S\left(z\right))}^2+2\int z(\frac{1}{2}-S\left(z\right))S{}^{\prime }\left(z\right)dz$$$$\int z(\frac{1}{2}-S\left(z\right))S{}^{\prime }\left(z\right)dz=$$$$=\int zsin\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-S\left(z\right))dz=$$$$=-\frac{1}{\pi }\int (\frac{1}{2}-S\left(z\right))d\left(cos\left(\frac{\pi z^2}{2}\right)\right)=$$$$=-\frac{1}{\pi }cos\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-S\left(z\right))-\frac{1}{\pi }\int cos\left(\frac{\pi z^2}{2}\right)S{}^{\prime }\left(z\right)dz$$$$\int cos\left(\frac{\pi z^2}{2}\right)S{}^{\prime }\left(z\right)dz=$$$$=\int cos\left(\frac{\pi z^2}{2}\right)sin\left(\frac{\pi z^2}{2}\right)dz=$$$$=\frac{1}{2\sqrt{2}}\int sin\left(\frac{\pi {\left(\sqrt{2}z\right)}^2}{2}\right)d\left(\sqrt{2}z\right)=$$$$=\frac{S\left(\sqrt{2}z\right)}{2\sqrt{2}}$$$$\Rightarrow \int z(\frac{1}{2}-S\left(z\right))S{}^{\prime }\left(z\right)dz=$$$$=-\frac{1}{\pi }cos\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-S\left(z\right))-\frac{S\left(\sqrt{2}z\right)}{2\sqrt{2}\pi }$$$$\Rightarrow G\left(z\right)=\int {(\frac{1}{2}-S\left(z\right))}^{2}dz=$$$$=z{(\frac{1}{2}-S\left(z\right))}^2-\frac{2}{\pi }cos\left(\frac{\pi z^2}{2}\right)(\frac{1}{2}-S\left(z\right))-\frac{S\left(\sqrt{2}z\right)}{\sqrt{2}\pi }$$$$G\left(0\right)=-\frac{1}{\pi }$$$$G\left(\mathrm{\infty }\right)=-\frac{1}{2\sqrt{2}\pi }$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{(\frac{1}{2}-S\left(z\right))}^{2}dz=\frac{1}{\pi }(1-\frac{1}{2\sqrt{2}})=\frac{4-\sqrt{2}}{4\pi }$$

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