MHB What has been done here to simplify the integration

[nacho-man]

Please refer to the attached image.

My lecturer seems to have re-written $e^{-st}$ as $\frac{d}{dt}e^{-st}$ and taken out the $(-\frac{1}{s}$ which I do see is equivalent, but i am unsure how he goes from there onwards.

Although, e is the derivative of itself. how does he split the integral into two, despite there being a multiplication of the terms?

 

[chisigma]

Please refer to the attached image.

My lecturer seems to have re-written $e^{-st}$ as $\frac{d}{dt}e^{-st}$ and taken out the $(-\frac{1}{s}$ which I do see is equivalent, but i am unsure how he goes from there onwards.

Although, e is the derivative of itself. how does he split the integral into two, despite there being a multiplication of the terms?

Your lecturer has applied the 'integration by parts' rule...

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

hah. thanks..
that was mildly embarrassing.

i didn't closely to see if he had done so. when he was talking in the lecture he mentioned that there was a 'trick' way to integrate this w/o integrating by parts or something along those lines.

 

[Ackbach]

hah. thanks..
that was mildly embarrassing.

i didn't closely to see if he had done so. when he was talking in the lecture he mentioned that there was a 'trick' way to integrate this w/o integrating by parts or something along those lines.

I would ask him again what it was he said. By-parts is certainly the standard way to integrate this, and it's not all that difficult, once you know how. I suppose you could set up tabular integration, but that's just a unified way of keeping track of by-parts. It wouldn't be worth it for only one application of by-parts.

 

[HallsofIvy]

The first step depends upon the fact that \frac{d(e^{-st})}{dt}= -s e^{-st} so that, dividing both sides by -s, e^{-st}= -\frac{1}{s}\frac{d(e^{-st})}{dt}.

Of course, since "s" is independent of the integration variable, t, we can take -\frac{1}{s} out of the integral.