Question

how to do derivative of transforms thingy? for example:

show that \(\Large\mathcal L \lbrace t^{1/2} \rbrace = \frac{1}{2s} \left(\frac{\pi}{s}\right) ^{1/2}, \quad \quad s>0\)

Answer 1

for the record...i have no idea

Answer 2

Can you use Integration By Parts here:

\[\large \mathsf{\int\limits\limits_{0}^{\infty} (t^\frac{1}{2}) \cdot (e^{-st})dt}\]

Answer 3

uhmm i have to use derivative of transforms instead of the usual..

Answer 4

By this we can find the derivative but I will give you directly how to derivative first time:

\[\large \mathcal{L[f'(t)] = s \cdot L[f(t)] - }\textbf {f(0)}\]

Answer 5

??

Answer 6

Solve for RHS here..

Answer 7

i have no idea or background knowledge on derivative of transforms...

Answer 8

Find:

\(\mathcal{L{f(t)}}\) first and then multiply it by s..

Answer 9

\[\large \mathcal{L(f(t)}) = \int\limits\limits _{0} ^{ \infty} (t^{\frac{1}{2}} \cdot e^{-st}) dt\]

Answer 10

\[\mathcal L\{f(t)\}=\int_0^∞f(t)e^{-st}\text dt\]

\[\mathcal L\{t^{1/2}\}=\int_0^∞t^{1/2}e^{-st}\text dt\]
\[=e^{-s}\int_0^∞t^{1/2}e^{t}\text dt\]\[=e^{-s}\Gamma \left( \frac{3}{2}\right)\]\[=e^{-s} \frac12\Gamma\left(\frac 12\right)\]\[=e^{-s}\frac {\sqrt{\pi}}{2}\]

ive dones something wrong, but im sure this will help a bit

Answer 11

\[\large e^{-st} \ne e^{-s} \cdot e^{t}\]

Answer 12

..i still ahve no idea what you are saying o.O

Answer 13

i often try to do that even thought it is illegal

Answer 14

I mean how you separate \(e^{-st}\) from the integral ??

Answer 15

can someone tell me how to derive transforms first?

Answer 16

If it was:

\[\large e^{-s +t} = e^{-s} \cdot e^{t}\]

Then you can do this..

Answer 17

what should i have done after this

\[\mathcal L\{t^{1/2}\}=\int_0^∞t^{1/2}e^{-st}\text dt\]\[\qquad\qquad=\]

Answer 18

We have to use here Integration By Parts..

Answer 19

maybe i should go and give you guys some space to discuss what you're discussing...

Answer 20

Can you check one thing @lgbasallote

Answer 21

i had no idea what any of you were saying...how can i check anything o.O

Answer 22

espansion? series? what??!

Answer 23

i dont know i havent studied laplace in a while,
i dont think series expansion is a good idea, maybe completing the square ,

Answer 24

\[\large \int\limits\limits _{0} ^{\infty}(t^{\frac{-1}{2}} \cdot e^{-st}) dt = (\frac{\pi}{s})^{\frac{1}{2}}\]

Here now @mukushla will help us..

Answer 25

what's going on here?? o.O

Answer 26

someone wanna tell me something =_=

Answer 27

you all have your own worlds -_-

Answer 28

@mukushla can you prove that I have written the integral above in my just last post..??

Answer 29

NOTE: im asking about derivative of transforms....

Answer 30

I am doing the same I guess..

Answer 31

i just need a demo on taking derivative of transform not some fancy tricks to display your awesomeness =_=

Answer 32

I am not awesome in laplace I was just helping you a bit in proving what you have written above..

Answer 33

yeah i know. i just love being sarcastic

Answer 34

i have no idea how to take the derivative of transform...i dont get what you linked me too..i hate links...

Answer 35

I also hate the links..

Answer 36

so how do you take derivativee of transforms?

Answer 37

i'll just try a simpler example...

Answer 38

well let

\[F(s)=\int\limits_{0}^{\infty} f(t,s) dt\] then u can write
\[\frac{dF}{ds}=\int\limits_{0}^{\infty} \frac{\partial f}{\partial s} dt\]
if second integral is convergence...

Answer 39

ugh..not good with letters =_= i'll just ask a simpler question and maybe i can get a demo there

Answer 40

huh??

Out of my knowledge..

Answer 41

u just clarify us....derivative of transforms or transform of derivatives?

Answer 42

\[\mathcal L\{t^{1/2}\}=\int\limits\limits_0^∞ t^{1/2}e^{-st}\text dt \implies \left| - \frac{t^{\frac{1}{2}} \cdot e^{-st}}{s} \right|_0^{\infty} + \frac{1}{2s} \int\limits\limits _{0}^{\infty} t^{\frac{-1}{2}} \cdot e^{-st}dt\]

.\[\large \implies 0 + \frac{1}{2s} \mathcal{L[t^{\frac{-1}{2}}]} \implies \frac{1}{2s} \frac{\Gamma (\frac{1}{2})}{s^{\frac{1}{2}}}\]

\[\Gamma (\frac{1}{2}) = \sqrt{\pi}\]

\[\large \mathcal{L[t^{\frac{1}{2}}]} = \frac{1}{2s} \sqrt{\frac{\pi}{s}}, \; \; s > 0\]

Answer 43

found the answer already...@sami-21 taught me gamma functions \m/