Idriss Olivier Bado
Idriss Bado
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integraleSep 7, 2026

Evaluation d une integrale

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idriss-bado
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Pour a,b,c>0a,b,c>0, calculer

I(a,b,c)=∫0+∞ln⁡(a2t)b2t2+c2 dt. I(a,b,c)=\int_0^{+\infty}\frac{\ln(a^2t)}{b^2t^2+c^2}\,dt.

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idriss-bado
Research member
Sep 7, 2026

Pour a,b,c>0a,b,c>0, posons

I(a,b,c)=∫0+∞ln⁡(a2t)b2t2+c2 dt. I(a,b,c)=\int_0^{+\infty}\frac{\ln(a^2t)}{b^2t^2+c^2}\,dt.

Effectuons le changement de variable

x=btc,t=cbx,dt=cb dx. x=\frac{bt}{c}, \qquad t=\frac{c}{b}x, \qquad dt=\frac{c}{b}\,dx.

Alors

I(a,b,c)=1bc∫0+∞ln⁡ ⁣(a2cbx)1+x2 dx. I(a,b,c) = \frac{1}{bc} \int_0^{+\infty} \frac{\ln\!\left(\frac{a^2c}{b}x\right)}{1+x^2}\,dx.

Ainsi

I(a,b,c)=1bc[ln⁡ ⁣(a2cb)∫0+∞dx1+x2+∫0+∞ln⁡x1+x2 dx]. I(a,b,c) = \frac{1}{bc} \left[ \ln\!\left(\frac{a^2c}{b}\right) \int_0^{+\infty}\frac{dx}{1+x^2} + \int_0^{+\infty}\frac{\ln x}{1+x^2}\,dx \right].

Or

∫0+∞dx1+x2=π2, \int_0^{+\infty}\frac{dx}{1+x^2}=\frac{\pi}{2},

et, par le changement de variable x↦1/xx\mapsto 1/x,

∫0+∞ln⁡x1+x2 dx=0. \int_0^{+\infty}\frac{\ln x}{1+x^2}\,dx=0.

Par conséquent,

I(a,b,c)=π2bcln⁡ ⁣(a2cb). \boxed{ I(a,b,c) = \frac{\pi}{2bc} \ln\!\left(\frac{a^2c}{b}\right) }.

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