Calculate Laplace Transform{t}

**MSUMathStdnt** · August 6th 2011, 09:55 AM

I need help completing (or have already made a mistake) calculating the Laplace Transform of g(t)=t. My work (1st equal: def of Lapl xform; 2nd equal: integration by parts):

$L \{ t \} = \int_0^{\infty} t e^{-st} dt = \frac{t e^{-st}}{-s} - \int_0^{\infty} \frac{e^{-st}}{-s} dt = \left[ \frac{t e^{-st}}{-s} - \frac{e^{-st}}{s^2} \right]_0^\infty$

$= lim_{b \rightarrow \infty} \left[ \frac{b e^{-sb}}{-s} - \frac{e^{-sb}}{s^2} \right] - \left[ \frac{0 \cdot e^{-s0}}{-s} - \frac{e^{-s \cdot0}}{s^2} \right] = \left[ \frac{\infty \cdot e^{- \infty}}{-s} - 0 \right] - \left[ 0 - \frac1{s^2} \right]$

$= \frac{\infty \cdot 0}{-s} + \frac1{s^2}$

P.S. How do you make an euler's constant "e" so it doesn't look like a variable e?

**Ackbeet** · August 6th 2011, 10:28 AM

I have the following:

$\mathcal{L}[t]=\int_{0}^{\infty}t\,e^{-st}\,dt=\frac{t\,e^{-st}}{-s}\Bigg|_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{-st}}{-s}\,dt=\left[\lim_{t\to\infty}\frac{t\,e^{-st}}{-s}\right]-\frac{e^{-st}}{s^{2}}\Bigg|_{0}^{\infty}.$

To evaluate the remaining limits, you must assume that $\text{Re}(s)\ge 0.$ If that's true, then you can use L'Hopital's Rule to finish them off.

**MSUMathStdnt** · August 6th 2011, 11:24 AM

Can't use l'Hospital's rule. The first derivative of the bottom of $\frac{te^{-st}}{-s}$ is zero.

**Ackbeet** · August 6th 2011, 11:27 AM

Originally Posted by MSUMathStdnt

Can't use l'Hospital's rule. The first derivative of the bottom of $\frac{te^{-st}}{-s}$ is zero.

Ah, but you can. Put the exponential in the denominator and flip the sign of the exponent.

**MSUMathStdnt** · August 6th 2011, 11:31 AM

Originally Posted by Ackbeet

Ah, but you can. Put the exponential in the denominator and flip the sign of the exponent.

Clearly, I must review l'Hospital's. Thanks.

**MSUMathStdnt** · August 19th 2011, 10:31 PM

Originally Posted by Ackbeet

I have the following:

$\mathcal{L}[t]=\int_{0}^{\infty}t\,e^{-st}\,dt=\frac{t\,e^{-st}}{-s}\Bigg|_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{-st}}{-s}\,dt=\left[\lim_{t\to\infty}\frac{t\,e^{-st}}{-s}\right]-\frac{e^{-st}}{s^{2}}\Bigg|_{0}^{\infty}.$

To evaluate the remaining limits, you must assume that $\text{Re}(s)\ge 0.$ If that's true, then you can use L'Hopital's Rule to finish them off.

Just cleaning up some old posts. I think the answer to the limit is

$\lim_{t\to\infty}\frac{t\,e^{-st}}{-s} = \lim_{t\to\infty}\frac{-t}{s\,e^{st}} = \text{l'Hospital's rule} = \lim_{t\to\infty}\frac{-1}{s^2\,e^{st}} = 0$ .

**Ackbeet** · August 20th 2011, 09:03 AM

Originally Posted by MSUMathStdnt

Just cleaning up some old posts. I think the answer to the limit is

$\lim_{t\to\infty}\frac{t\,e^{-st}}{-s} = \lim_{t\to\infty}\frac{-t}{s\,e^{st}} = \text{l'Hospital's rule} = \lim_{t\to\infty}\frac{-1}{s^2\,e^{st}} = 0$ .

Looks correct to me.

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