Question

Can someone please show me how to integrate (from 0 to infinity) (1-e^-t)(e^(-st)). Wolfram Alpha shows ln(1+1/s) but it does not show the steps to prove it.

Answer 1

do you need help

Answer 2

can you post the link to wolfram so i can look at the integral

Answer 3

OH no I've got the problem written incorrectly! http://www.wolframalpha.com/input/?s=48&_=1416113858777&i=integrate+((1-e%5e-t)%2ft)(e%5e(-st))+from+0+to+infinity&fp=1&incTime=true

Answer 4

that is the problem I need help with! I know the answer should come out to be ln(1+1/s) but I'm unsure how to get it.

Answer 5

integration by parts should do right ?
by the way, this is the integral you need to evaluate for finding laplace transform of \(\large \rm \frac{1-e^{-t}}{t}\)

http://www.wolframalpha.com/input/?i=laplace+transform+%281-e%5E-t%29%2Ft

also notice \(\large \rm s\gt 0\) for the integral to converge

Answer 6

Yea I know It's the Laplace transform but I have to show all of the steps and I don't know if I'm screwing it up or what but I can't get integration by parts to work

Answer 7

@hartnn

Answer 8

so LT can't be used ?

Answer 9

i think it should be okay to use but there is no LT for 1/t and integration by parts doesnt seem to work

Answer 10

tried with this ?

Answer 11

i wonder whether Differentiation under integral sign method
helps to get the answer without using LT

Answer 12

nope, i am still messing with by parts but its hopeless really

Answer 13

http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

Answer 14

\[\begin{align} I(s) &= \int\limits_0^{\infty} \dfrac{1-e^{-t}}{t}e^{-st} \; dt\\~\\
I'(s)&= \int\limits_0^{\infty} \dfrac{\partial}{\partial s}\dfrac{1-e^{-t}}{t}e^{-st} \; dt \\~\\
&= \int\limits_0^{\infty} -s(1-e^{-t})e^{-st} \; dt \\~\\
&= s\int\limits_0^{\infty} -e^{-st} +e^{-t(s+1)} \; dt \\~\\
& = s\left[\dfrac{e^{-st}}{s} - \dfrac{e^{-t(s+1)}}{s+1}\right]_{t=0}^{t=\infty} \\~\\
&=-\dfrac{1}{s+1}
\end{align}\]

integrate both sides and thank @hartnn :)

Answer 15

small mistake :\[\begin{align} I(s) &= \int\limits_0^{\infty} \dfrac{1-e^{-t}}{t}e^{-st} \; dt\\~\\
I'(s)&= \int\limits_0^{\infty} \dfrac{\partial}{\partial s}\dfrac{1-e^{-t}}{t}e^{-st} \; dt \\~\\
&= \int\limits_0^{\infty} -(1-e^{-t})e^{-st} \; dt \\~\\
&= \int\limits_0^{\infty} -e^{-st} +e^{-t(s+1)} \; dt \\~\\
& = \left[\dfrac{e^{-st}}{s} - \dfrac{e^{-t(s+1)}}{s+1}\right]_{t=0}^{t=\infty} \\~\\
&=-\dfrac{1}{s}+\dfrac{1}{s+1}
\end{align}\]

Answer 16

http://math.stackexchange.com/questions/1024379/evaluate-int-limits-0-infty-dfrac1-e-tte-st-dt

Answer 17

is the final answer that above. Should there be an ln in there somewhere?

Answer 18

you have \[\large \rm I'(s) = -\dfrac{1}{s} + \dfrac{1}{s+1}\]

integrate both sides for\(\large \rm I(s)\)

Answer 19

ahhhhhh ok thanks!!!!!

Answer 20

np:) go thru that MSE link if u have time... it has half dozen different methods for evaluating this integral