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Question Number 141649 by mnjuly1970 last updated on 21/May/21

$$.......advancedcalculus......$$$$provethat-::$$$$\varphi :={\int }_0^{\mathrm{\infty }}\frac{cos\left(2\pi x^2\right)}{{cosh}^2\left(\pi x\right)}dx=\frac{1}{4}✓$$

Answered by ArielVyny last updated on 22/May/21

$$$$$$\varphi ={\int }_0^{\propto }\frac{cos\left(2\pi x^2\right)+1}{{cosh}^2\left(\pi x\right)}dx-{\int }_0^{\propto }\frac{1}{{cosh}^2\left(\pi x\right)}dx$$$$\varphi ={\varphi }_1-{\varphi }_2$$$${\varphi }_2={\int }_0^{\propto }\frac{1}{{cosh}^2\left(\pi x\right)}dx$$$$\pi x=t\pi dx=dt$$$${\varphi }_2=\frac{1}{\pi }{\int }_0^{\propto }\frac{1}{{cosh}^2\left(t\right)}dt=\frac{1}{\pi }$$$${\varphi }_1={\int }_0^{\propto }\frac{cos\left(2\pi x^2\right)+1}{{cosh}^2\left(\pi x\right)}dx$$$${cos}^2\left(\pi x^2\right)=\frac{1+cos\left(2\pi x^2\right)}{2}$$$${\varphi }_1={\int }_0^{\propto }\frac{2{cos}^2\left(\pi x^2\right)}{{cosh}^2\left(\pi x\right)}dx$$$${\varphi }_1=2{\int }_0^{\propto }{\left(\frac{cos\left(\pi x^2\right)}{cosh\left(\pi x\right)}\right)}^{2}dx$$$${\varphi }_1=2{\int }_0^{\propto }{\left(\frac{{(-1)}^{x^2}}{cosh\left(\pi x\right)}\right)}^{2}dx$$$${\varphi }_1=2{\int }_0^{\propto }{\left(\frac{1}{cosh\left(\pi x\right)}\right)}^{2}dx$$$${\varphi }_1=2{\int }_0^{\propto }\frac{1}{{cosh}^2(\pi x}dx$$$$\pi x=t.\pi dx=dt$$$${\varphi }_1=2\frac{1}{\pi }{\int }_0^{\propto }\frac{1}{{cosh}^2\left(t\right)}dt$$$${\varphi }_1=\frac{2}{\pi }$$$$\varphi =\frac{1}{\pi }$$$$checkintheanswerbecauseidonothaveyourresult.$$

Commented by mnjuly1970 last updated on 22/May/21

$$thankyousomuch$$$$iwillrechck...$$

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