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Question Number 65878 by ~ À ® @ 237 ~ last updated on 05/Aug/19

$$Calculate{lim}_{a->\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^a}$$

Commented by mathmax by abdo last updated on 05/Aug/19

$$changement{x}^a=tgivex=t^{\frac{1}{a}}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^a}={\int }_0^{\mathrm{\infty }}\frac{1}{1+t}\frac{1}{a}{t}^{\frac{1}{a}-1}dt$$$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{a}-1}}{1+t}dtsofora>1weget0<\frac{1}{a}<1\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^a}=\frac{1}{a}\frac{\pi }{sin\left(\frac{\pi }{a}\right)}\Rightarrow {lim}_{a\to +\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^a}={lim}_{a\to +\mathrm{\infty }}\frac{\frac{\pi }{a}}{sin\left(\frac{\pi }{a}\right)}$$$$={lim}_{t\to 0}\frac{t}{sint}=1(putt=\frac{\pi }{a})finally$$$${lim}_{a\to +\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+x^a}=1$$

Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19

$$thankyousir!butatthesecondline,{it}^{\prime }snotclear$$

Commented by mathmax by abdo last updated on 05/Aug/19

$$x=t^{\frac{1}{a}}\Rightarrow dx=\frac{1}{a}{t}^{\frac{1}{a}-1}dt\dots .$$

Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^a}.As\frac{1}{1+x^a}<\frac{1}{x^a}\Rightarrow f\left(a\right)isreal$$$$changeu=\frac{1}{1+x^a}thenx={(\frac{1}{u}-1)}^{\frac{1}{a}}anddx=\frac{1}{a}.\frac{-1}{u^2}.{(\frac{1}{u}-1)}^{\frac{1}{a}-1}du$$$$Then$$$$f\left(a\right)=\frac{1}{a}{\int }_0^{1}u.\frac{1}{u^2}.{\left(\frac{1-u}{u}\right)}^{\frac{1}{a}-1}du$$$$=\frac{1}{a}{\int }_0^1{u}^{\frac{-1}{a}}{(1-u)}^{\frac{1}{a}-1}du$$$$=\frac{1}{a}{\int }_0^1{u}^{1-\frac{1}{a}-1}{(1-u)}^{\frac{1}{a}-1}du=\frac{1}{a}B(1-\frac{1}{a},\frac{1}{a})=\frac{1}{a}.\frac{\mathrm{\Gamma }(1-\frac{1}{a})\mathrm{\Gamma }\left(\frac{1}{a}\right)}{\mathrm{\Gamma }\left(1\right)}$$$$As\mathrm{\forall }z\mathrm{\Gamma }(1-z)\mathrm{\Gamma }\left(z\right)=\frac{\pi }{sin\left(\pi z\right)}wefinallyget$$$$f\left(a\right)=\frac{\frac{\pi }{a}}{sin\left(\frac{\pi }{a}\right)}theneasily{lim}_{a->\mathrm{\infty }}f\left(a\right)=1(bychangingb=\frac{\pi }{a})$$

Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19

$$As\frac{1}{1+x^a}<\frac{1}{x^a}\Rightarrow f\left(a\right)isrealwhena>1$$

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