Laplace transformation giving $zero$

Question

I've tried a lot to answer this $$\int_0^{+\infty}\frac{e^{-3t}-e^{-6t}}{t}$$ but is only giving $0$ like answer

So, I tried for Laplace transformation, but it didnt give me a good answer (with $(-1)!$), and when i tried with normal intagration, i found it $zero$.

But, i know the right answer is $ln(2)$ (acoording to the question)

Can someone help? :)

Answer 1

Rewrite the integral like so

$$\int_0^\infty \frac{e^{-3t}-e^{-6t}}{t}\:dt = \int_0^\infty \int_3^6 e^{-st}\:ds\:dt$$

then swap the order of integration to get

$$\int_3^6 -\frac{e^{-st}}{s}\Biggr|_0^\infty \:ds = \int_3^6 \frac{ds}{s} = \ln(2)$$