Why $ \int_0^{\infty} du \, \frac{e^{-3 u} - e^{-4 u}}{u} = \int_0^{\infty} du \, \int_3^4 dt \, e^{-u t} \\ $?

Question

from this answer I could not see what is happening here:

$$ \int_0^{\infty} du \, \frac{e^{-3 u} - e^{-4 u}}{u} = \int_0^{\infty} du \, \int_3^4 dt \, e^{-u t} \\ $$

What technique of integration is being used here?

Answer 1

$$\int_3^4 e^{-ut}\,dt=\left.-\frac1ue^{-ut}\right|_3^4=-\frac1u(e^{-4u}-e^{-3u})=\frac{e^{3u}-e^{4u}}u$$

and now substitute the above in the right hand of your equality