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Question Number 171071 by mathocean1 last updated on 07/Jun/22

$$justifythat{\int }_0^{+\mathrm{\infty }}\frac{dt}{1+t^4}isconvergent.$$

Answered by aleks041103 last updated on 07/Jun/22

$${\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}={\int }_0^1\frac{dt}{1+t^4}+{\int }_1^{\mathrm{\infty }}\frac{dt}{1+t^4}$$$$1+t^4⩾1for0⩽t⩽1$$$$\Rightarrow \frac{1}{1+t^4}⩽1\Rightarrow {\int }_0^1\frac{dt}{1+t^4}<{\int }_0^{1}1dt=1$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}<1+{\int }_1^{\mathrm{\infty }}\frac{dt}{1+t^4}$$$$for{t}^4⩾t^{2}fort⩾1$$$$\Rightarrow \frac{1}{1+t^4}⩽\frac{1}{1+t^2}$$$$\Rightarrow {\int }_1^{\mathrm{\infty }}\frac{dt}{1+t^4}⩽{\int }_1^{\mathrm{\infty }}\frac{dt}{1+t^2}=arctg{\left(t\right)}_1^{\mathrm{\infty }}=\frac{\pi }{4}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}<1+\frac{\pi }{4}$$$$obv.$$$$\frac{1}{1+t^4}>0\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}>0$$$$\Rightarrow 0<{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}<1+\pi /4$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}converges$$

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