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# Integration

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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
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#1
+26096
+3

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

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

Alan  Jun 17, 2017
#2
+2

Excellent, thank you for explaining.

Guest Jun 17, 2017
#3
+90134
+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
#6
+26096
0

Your last step has the closing bracket in the wrong place Melody! It should be:

$$\frac{u}{s^2}(\ln u-1)+c$$

(The constant shouldn't be multiplied by u, a function of t).

This also affects Max's extension below (now corrected by Max).

Alan  Jun 18, 2017
edited by Alan  Jun 18, 2017
edited by Alan  Jun 18, 2017
#4
+6779
+1

But u = e^(-st) so:

$$\dfrac{u}{s^2}(\ln u-1)+k\\ =\dfrac{e^{-st}}{s^2}(-st-1)+k$$

MaxWong  Jun 18, 2017
edited by MaxWong  Jun 18, 2017
#5
+90134
+1

Yes of course. Thanks Max.

I still wnder where that extra step in the question could ever be useful :)

Melody  Jun 18, 2017

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