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What has been done here to simplify the integration

nacho

Active member
Sep 10, 2013
156
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?
 

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chisigma

Well-known member
Feb 13, 2012
1,704
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

Active member
Sep 10, 2013
156
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

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The first step depends upon the fact that [tex]\frac{d(e^{-st})}{dt}= -s e^{-st}[/tex] so that, dividing both sides by -s, [tex]e^{-st}= -\frac{1}{s}\frac{d(e^{-st})}{dt}[/tex].

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