Proof for this Laplace transform

In summary, the Laplace transform of the function t^n is given by L(t^n) = \frac{n!}{s^{n+1}}, which can be proved using induction.
  • #1
Reshma
749
6
Can someone help me out with the proof of the Laplace transform of the function tn?

I did have a go at this one.
[tex]L[t^n] = \int_0^{\infty} t^n e^{-st}dt[/tex]

[tex]= t^n\frac{-e^{-st}}{s}\vert_0^{\infty} + {1\over s}\int_0^{\infty} nt^{n-1}e^{-st}dt[/tex]

[tex]={n\over s}\int_0^{\infty} t^{n-1}e^{-st}dt[/tex]

[tex]={n\over s}L[t^{n-1}][/tex]

I am supposed to arrive at the result:
[tex]L[t^n] = \frac{(n+1)!}{s^{n+1}}[/tex]
 
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  • #2
Just an addendum, should I prove this using induction?
 
  • #3
[tex]L(t^{n}) = \int_{0}^{\infty} t^{n}e^{-st}dt[/tex]

Let [tex]v=st[/tex] to give [tex]L(t^{n}) = \frac{1}{s^{n+1}}\int_{0}^{\infty}v^{n}e^{-v}dv = \frac{1}{s^{n+1}}\Gamma(n+1)[/tex]

[tex]\Gamma(n+1) = n![/tex], giving the result [tex]L(t^{n}) = \frac{n!}{s^{n+1}}[/tex]. This isn't what you've got, but check by using n=1 to see whose correct.

[tex]L(t) = \int_{0}^{\infty}te^{-st}dt =\left[ \frac{te^{-st}}{-s}\right]_{0}^{\infty} - \int_{0}^{\infty}\frac{1}{-s}e^{-st}dt = \frac{1}{s^{2}}[/tex]

Hence [tex]L(t^{1}) = \frac{1!}{s^{2}}[/tex], so the formula you were trying to derive isn't correct.
 
Last edited:

Related to Proof for this Laplace transform

1. What is a Laplace transform and how is it used in science?

A Laplace transform is a mathematical tool that converts a function of time into a function of complex frequency. In science, it is commonly used to simplify and solve differential equations, which are often used to model physical systems.

2. Why is the Laplace transform considered useful in scientific research?

The Laplace transform has many advantages in scientific research, including its ability to transform complicated functions into simpler algebraic equations, the ability to handle non-continuous and discontinuous functions, and the ability to solve initial value problems in a straightforward manner.

3. How is the Laplace transform related to the Fourier transform?

The Laplace transform is a generalization of the Fourier transform, as it extends the domain from purely periodic signals to a wider class of signals. The Fourier transform can be seen as a special case of the Laplace transform, where the complex frequency is restricted to the imaginary axis.

4. Are there any limitations to using the Laplace transform?

While the Laplace transform is a powerful tool, it does have some limitations. It can only be applied to functions that are defined for all positive values of time, and it is not as widely used in engineering and physics as the Fourier transform. Additionally, the inverse Laplace transform can be difficult to compute for certain functions.

5. Can the Laplace transform be used in real-world applications?

Yes, the Laplace transform has many practical applications in engineering, physics, and other fields. It is commonly used in circuit analysis, control systems, signal processing, and solving differential equations in physics and engineering problems.

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