Bonjour
Solution postée
A+
voici la solution d'abdelbaki
Bonjour,
on pose t=ax , dt/t=dx/x ==> (\int_a^b) e^{t/a} dt/t = (\int_1^{b/a}) e^x dx/x
on pose t=b/x, dt/t =-dx/x ==> (\int_a^b) e^{b/t} dt/t = - (\int_{b/a}^1) e^x dx/x
==> I=0
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