Laplace Transform of t by Sine a t/Proof 2
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Theorem
Let $\sin$ denote the real sine function.
Let $\laptrans f$ denote the Laplace transform of a real function $f$.
Then:
- $\laptrans {t \sin a t} = \dfrac {2 a s} {\paren {s^2 + a^2}^2}$
Proof
We have:
\(\ds \laptrans {\cos a t}\) | \(=\) | \(\ds \int_0^\infty e^{-s t} \cos a t \rd t\) | Definition of Laplace Transform | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac s {s^2 + a^2}\) | Laplace Transform of Cosine |
Hence:
\(\ds \dfrac \d {\d a} \int_0^\infty e^{-s t} \cos a t \rd t\) | \(=\) | \(\ds \int_0^\infty e^{-s t} \paren {\map {\dfrac \partial {\partial a} } {\cos a t} } \rd t\) | Derivative of Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty e^{-s t} \paren {-t \sin a t} \rd t\) | Derivative of Cosine Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\laptrans {t \sin a t}\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\dfrac \d {\d a} } {\dfrac s {s^2 + a^2} }\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac {2 a s} {\paren {s^2 + a^2}^2}\) | Quotient Rule for Derivatives |
and the result follows.
$\blacksquare$
This article, or a section of it, needs explaining. In particular: I know we have that result Derivative of Integral somewhere, I just can't find it. Spiegel calls it Leibnitz's Rule (not mis-spelling) but I know we have it as something else. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Multiplication by Powers of $t$: $20 \ \text{(a)}$