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Does the Laplace Teansform of the function exist? 
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#1
Mar2213, 02:50 AM

P: 14

While reading Laplace transform in my book, I came across a problem. where it evaluates the laplace transform of
f(t)=1/√t => √(π/s) . But is the laplace transform of f(t) really exists? because I thought the function f(t) is not defined at t=0 and it tends to infinite. So How is it possible that the definite intgeration over [0,∞] can be evaluated for such function? 


#2
Mar2213, 09:25 PM

P: 581

Just because a function goes to infinity at the boundary of a region does not mean the definite integral over that region does not have a finite value. Consider
[itex]\int_0^4 \frac{1}{\sqrt{x}}dx = 2\sqrt{x}_0^4 =2\sqrt{4} = 4 [/itex] So for the example above, even though the function goes to infinity as x goes to zero, the antiderivative does not. 


#3
Mar2413, 09:31 AM

P: 14




#4
Mar2413, 09:58 AM

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Does the Laplace Teansform of the function exist?
For a mathematical proof of this, you need a course on real analysis (not a beginning "calculus" course that teaches you some "rules" for doing integration and differentiation, but without any proofs). 


#5
Mar2413, 10:03 AM

P: 446

A good way to think about it (unrigorously) is that the function, where it tends to infinity, is very narrow. Again being loose with terminology, if the function is infinitely high but infinitely narrow, the area might be finite.
Thats just an intuitive way to think. By the FTC [itex]\int_0^x f(t)dt=F(x)F(0)[/itex] we can also use limits, but for your example that isn't necessary. As long as the antiderivative is finite (and continuous on the domain) the area under the curve must be finite even though the curve tends to infinity 


#6
Mar2613, 12:33 PM

P: 14




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