Hi,

I am studying a math course and the following operation is performed on an integral but is not explained:

\(F(s)=\int{te^{-st}}dt=-{1\over{s}}\int{td(e^{-st}})\)

Could somebody explain what happened here?

Guest Jun 17, 2017

#1**+3 **

Let u = e^{-st} then du/dt = -se^{-st} or e^{-st}dt = -(1/s)du so we can write the integral as:

\(\int te^{-st}dt \rightarrow -\frac{1}{s}\int tdu \rightarrow -\frac{1}{s}\int td(e^{-st})\)

Alan
Jun 17, 2017

#3**+1 **

Thanks Alan,

I would not have got that either without you showing me.

I just want to finish it.

Let

\(u = e^{-st} \;\; then\;\; \\\frac{du}{dt} = -se^{-st} \\e^{-st}dt = \frac{-1}{s}du \)

so we can write the integral as:

\(\int te^{-st}dt \rightarrow -\frac{1}{s}\int tdu \rightarrow -\frac{1}{s}\int td(e^{-st})\)

\(u=e^{-st}\\ ln(u)=-st\\ t=\frac{-ln(u)}{s}\)

So

\(\frac{-1}{s}\int t\;du\\ =\frac{-1}{s}\int \frac{-ln(u)}{s}\;du\\ =\frac{1}{s^2}\int 1*ln(u)\;du\\ \qquad \text{integrate by parts}\\ \qquad \frac{da}{du}=1\quad b=ln(u)\\ \qquad a=u\quad \frac{db}{du}=\frac{1}{u}\\ =\frac{1}{s^2}\left( [ab]-\int a*\frac{db}{du}\:du\right)\\ =\frac{1}{s^2}\left( [uln(u)]-\int u*\frac{1}{u}\:du\right)\\ =\frac{1}{s^2}\left( [uln(u)]-\int 1\:du\right)\\ =\frac{1}{s^2}\left( [uln(u)]-u\right)+c\\ =\frac{u}{s^2}\left( ln(u)-1+k \right)\)

Why would you ever want to present it with the extra step in the question?

Melody
Jun 17, 2017