# What has been done here to simplify the integration

#### nacho

##### Active member
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
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
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
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
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.