Exercice
Étude de convergence **
15 mars 2024 22:26
— Par Michel Quercia
Étudier la convergence des intégrales suivantes :
\(\int _{t=-\infty }^{+\infty } \dfrac{d t}{e^t+t^2 e^{-t}}\)
\(\int _{t=1}^{+\infty } \dfrac{e^{\sin t}}t\,d t\)
\(\int _{t=0}^1 \dfrac{t^\alpha -1}{\ln t}\,d t\)
\(\int _{t=e^2 }^{+\infty } \dfrac{d t}{t(\ln t)(\ln\ln t)}\)
\(\int _{t=0}^{+\infty } \ln\Bigl( \dfrac{1+t^2 }{1+t^3}\Bigr)\,d t\)
\(\int _{t=0}^{+\infty } \Bigl(2+(t+3)\ln\bigl(\dfrac{t+2}{t+4}\bigr)\Bigr)d t\)
\(\int _{t=0}^{+\infty } \dfrac{t\ln t}{(1+t^2 )^\alpha }\,d t\)
\(\int _{t=0}^1 \dfrac{d t}{1-\sqrt t}\)
\(\int _{t=0}^{+\infty } \dfrac{(t+1)^\alpha -t^\alpha }{t^\beta }\,d t\)
\(\int _{t=0}^{+\infty } \sin(t^2 )\,d t\)
\(\int _{t=0}^1 \dfrac{d t}{\arccos t}\)
\(\int _{t=0}^{+\infty } \dfrac{\ln(\arctan t)}{t^\alpha }\,d t\)
\(\int _{t=1}^{+\infty } \dfrac{\ln(1+1/t)\,d t}{(t^2 -1)^\alpha }\)
\(\int _{t=0}^1 \dfrac{|\ln t|^\beta }{(1-t)^\alpha } \,d t\)
\(\int _{t=0}^{+\infty } t^\alpha \bigl(1-e^{-1/\sqrt t}\,\bigr)\,d t\)
\(\int _{t=0}^1 \sin(1/t)e^{-1/t}t^{-k}\,d t\)