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?
Let u = e-st then du/dt = -se-st or e-stdt = -(1/s)du so we can write the integral as:
\(\int te^{-st}dt \rightarrow -\frac{1}{s}\int tdu \rightarrow -\frac{1}{s}\int td(e^{-st})\)
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?