Laplace Transform: Integral of x from 0 to ∞

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SUMMARY

The discussion centers on the evaluation of the improper integral \(\int_{0}^{\infty} x \, dx\) and its relation to the Laplace Transform. The integral evaluates to \(\frac{1}{2}\) when considering limits, but the participants question how the upper limit of infinity is transformed into 1 during evaluation. Additionally, there is confusion regarding the connection between this integral and the Laplace Transform, particularly the necessity of including the exponential term \(e^{-xt}\) in the context of Laplace Transforms.

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catcherintherye
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\int_{0}^{\infty} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}
 
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always check your post AFTER you send it with latex code!
 
catcherintherye said:
\int_{0}^{\infty} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}

Is there a question here? My question is 'how did the \infty in the integral become 1 in the evaluation?

Also, in what sense does this have anything to do with a "Laplace transform"? could you have forgotten an e^{-xt}?
 
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