Today ' S Calculation Of Integrals - 4

Hey Ricky ... I might be wrong too.

Btw ..... what's the answer?????

Or, taking the given integral as a function of \alpha:
I(\alpha)=\int_0^\infty \dfrac{e^{-\alpha x}-e^{-\beta x}}{x}\ \mathrm dx
Differentiating w.r.t. \alpha, we get
\dfrac{\mathrm I(\alpha)}{\mathrm d\alpha}=\int_0^\infty -e^{-\alpha x}\ \mathrm dx =-\dfrac{1}{\alpha}
Hence,
I(\alpha)=-\ln \alpha + C
Obviously, I(\beta)=0, so that C=\ln \beta. Hence, we get
I(\alpha) =\ln \dfrac{\beta}{\alpha}