Question

Compute
\[
\int_{0}^{\infty} \frac{e^{-bx} - e^{-ax}}{x} \, dx
\]

Answer 1

Using Laplace Transform would be one of the easiest way!
I'll wait for any other easy method :)

Answer 2

My hunch is to undo the "integral" inside and then switch the order of integration. One sec...

Answer 3

So since we can look at this integral as being inside since they're equal: \[\Large \int\limits_b^a e^{-yx}dy=\frac{-1}{x}e^{-yx}|_b^a=\frac{e^{-bx}-e^{-ax}}{x}\] The integral is really: \[\Large \int\limits_0^\infty \int\limits_b^a e^{-yx}dy dx\] Switch the order of integration: \[\Large \int\limits_b^a \int\limits_0^\infty e^{-yx}dxdy= \int\limits_b^a \frac{-1}{y} \left( \frac{1}{e^\infty} - \frac{1}{e^0} \right) dy \\ \Large \int\limits_b^a \frac{1}{y}dy=\ln \frac{a}{b}\]

Answer 4

@Sambhavvinaykya That's not right since exponents will look like this: \[\Large e^x(e^{-a}-e^{-b}) =e^{-a+x}-e^{-b+x}\]

Answer 5

ohh sorry

Answer 6

That's fine, I was just trying to help you out. =D

Answer 7

thanks