$$$$

Exercices sur les intégrales impropres

Calculer, si possible:

• $I=\int_1^\infty\frac{a^x}{1+a^{2x}}dx$ avec $a\gt0$

  $$I=\int_1^\infty\frac{a^x}{1+a^{2x}}dx \quad a\gt0\n t=a^x\n \ln t=x\ln a\n x=\frac{\ln t}{\ln a}\quad\Rightarrow\quad dx=\frac1{\ln a}\frac{dt}{t}$$

  Disjonction de cas :

  • si $a\gt1$

    $$I = \int_a^0\frac{t}{1+t^2}\frac1{\ln a}\frac{dt}{t}\n I=\frac1{\ln a}\int_a^0\frac1{1+t^2}dt\n I = \frac1{\ln a} \cembrace{\arctg t}_{a}^{0} I = \frac1{\ln a}\pembrace{\frac\pi2-\arctg a}$$

  • si $a=1$

    $$I = \int_1^\infty\frac{1^x}{1+1^{2x}}\n I = \int_1^\infty\frac12dx$$

  • si $a\lt1$

    $$I=-\int_0^a\frac1{1+t^2}\frac{dt}{\ln a}\n I=\frac{-1}{\ln a}\cembrace{\arctg t}_0^a\n I=\frac{-1}{\ln a}\cembrace{\arctg a-\arctg 0}\n I = -\frac{\arctg a}{\ln a}$$

• $I_n=\int_0^\infty t^ne^{-t^2}$

  $$f(t)=t^ne^{-t^2}\text{ est continue sur }\R\n \embrace{ &\lim_{t\rightarrow\infty}t^ne^{-t^2}=0\\ &\lim_{t\rightarrow\infty}t^ne^{-t}=0 }{|}{.}\n$$

  1. $$\lim_{n\rightarrow\infty}t^{n+2}t^{-2}=0\n \exist A\gt0\n t^{n+2}t^{-2} \lt 1 \Leftrightarrow t^nt^{-2}\lt\frac1{t^2}\n \Rightarrow I_n \le \underbrace{\int_0^{\infty}\frac1{t^2}dt}_{CVG}\quad\boxed{CVG}\n\n I_0=\int_0^\infty e^{-t^2}dt=\frac{\sqrt{\pi}}2\n I_1=\int_0^\infty te^{-t^2}dt=\cembrace{-\frac12e^{-t^2}}_{0}^{\infty}=\frac12\n\n I_n=\int_°^\infty t^ne^{-t^2}dt\n I_n=\int_°^\infty t^{n-1}*te^{-t^2}dt\n \embrace{\ &u=t^{n-1}\quad u'=(n-1)t^{n-2}\\ &v'=te^{-t^2}\quad v=-\frac12e^{-t^2}}{|}{.}\n I_n=\underbrace{\cembrace{t^{n-1}*(-\frac12e^{-t^2})}_{0}^{\infty}}_{=0}+\int_0^\infty\frac12e^{-t^2}(n-1)t^{n-2}dt\n I_n=\frac12(n-1)\int_0^\infty t^{n-2}e^{-t^2}dt\n I_n=\frac{n-1}2I_{n-2}\n\n \begin{split} &I_2=\frac12*I_0=\frac{\sqrt{\pi}}4\n &I_3=1*I_1=\frac12\n &I_4=\frac32*I_2=\frac{3\sqrt{\pi}}8\n &I_5=2*I_3=1\n &I_6=\frac52*\frac{3\sqrt{\pi}}8\n &I_7=3*I_5=3\n &I_9=4*I_7=12\n \text{donc : }\\ &I_{2p+1} = f_1(p)*I_1\n &I_{2p}=f_2(p)*I_0 \end{split}$$

• $I=\int_0^\infty x\cos^4x_ dx$

  ParseError: KaTeX parse error: Undefined control sequence: \bembrace at position 51: … \int_0^\infty \̲b̲e̲m̲b̲r̲a̲c̲e̲{x\cos^4x}dx\le…

• $I=\int_0^\infty \frac{x^n}{1+x^2}dx$

  $$I=\int_0^\infty \frac{x^n}{1+x^2}dx\n \embrace{ &u=x^2\quad\quad\rightarrow u'=nx^{n-1}\\ &v'=\frac1{1+x^2}\rightarrow v=\arctg x }{|}{.}\n I_n=\underbrace{\cembrace{x^n\arctg x}_{0}^{\infty}}_\infty\n I_n\quad n\le2\quad \boxed{DVG}\n\n I_0=\int_0^\infty \frac1{1+x^2}dx\n I_0=\cembrace{\arctg x}_{0}^{\infty}\n I_0=\frac\pi2\n\n I_1:\quad \frac{x}{1+x^2}\xrightarrow[x\rightarrow\infty]{}\frac1x\quad\Rightarrow\boxed{DVG}$$