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Does the Laplace Teansform of the function exist?

by rudra
Tags: exist, function, laplace, teansform
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rudra
#1
Mar22-13, 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?
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MisterX
#2
Mar22-13, 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 anti-derivative does not.
rudra
#3
Mar24-13, 09:31 AM
P: 14
Quote Quote by MisterX View Post
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 anti-derivative does not.
To my understanding, integral means the area under the curve. If fn. tends to infinite at some region, the area should also be infinite under this curve. Please clarify your point.

AlephZero
#4
Mar24-13, 09:58 AM
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Does the Laplace Teansform of the function exist?

Quote Quote by rudra View Post
To my understanding, integral means the area under the curve.
That is correct...
If fn. tends to infinite at some region, the area should also be infinite under this curve.
... but that is not necessarily true. Intuitively, if a function goes to infinity "fast enough", the area under the curve can still be finite.

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).
DrewD
#5
Mar24-13, 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
rudra
#6
Mar26-13, 12:33 PM
P: 14
Quote Quote by DrewD View Post
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
Thank you guys for the clarification. It helped a lot.


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