- Thread starter
- #1

- Thread starter nacho
- Start date

- Thread starter
- #1

- Feb 13, 2012

- 1,704

You can use the fact that for n positive integer is...How do you find the laplace transform of this without expanding it?

$ L(t^2+1)^2 $

$\displaystyle \int_{0}^{\infty} t^{n}\ e^{- s\ t}\ dt = \frac{n!}{s^{n+1}}\ (1)$

... expanding $\displaystyle (1 + t^{2})^{2}$ in powers of t...

Kind regards

$\chi$ $\sigma$

I didn't realize immediately that it was requested 'without expanding' ... very sorry!...

- Feb 13, 2012

- 1,704

The possible solution is the use of a 'forgotten formula' ['forgotten' in the sense that it is neglected from most of the Complex Analysis 'Holybooks'...] according to which if You have two L-transformable functions $f_{1}(t)$ and $f_{2} (t)$ with L-transforms $F_{1} (s)$ and $F_{2} (s)$ and abscissas of convergence $\sigma_{1}$ and $\sigma_{2}$, then the L-transform of the product is given by the integral...How do you find the laplace transform of this without expanding it?

$ L(t^2+1)^2 $

$\displaystyle \mathcal {L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F_{1} (z)\ F_{2}(s - z)\ dz\ (1)$

... where $\displaystyle \sigma_{1} < \gamma < \sigma - \sigma_{2},\ \sigma> \sigma_{1} + \sigma_{2}$. In Your case is $\displaystyle f_{1}(t) = f_{2} (t) = 1 + t^{2} \implies F_{1}(s) = F_{2}(s) = \frac{1}{s} + \frac{2}{s^{3}},\ \sigma_{1}=\sigma_{2}=0$. May be it exists a more comfortable way but till now I didnn't succed to find it!...

Kind regards

$\chi$ $\sigma$

- Moderator
- #4

- Jan 26, 2012

- 995

To compute this without expanding requires a little trick (especially if you're allowed to do this without using the integral definition of $\mathcal{L}\{f(t)\}$).How do you find the laplace transform of this without expanding it?

$ L(t^2+1)^2 $

Let $f(t) = (t^2+1)^2$. We note that $f(0)=1$.

Thus, $f^{\prime}(t) = 4t(t^2+1)$ and hence $f^{\prime}(0) = 0$.

Finally, we see that $f^{\prime\prime}(t) = 4(t^2+1) + 8t^2 = 12t^2+4$.

Why did we compute all these derivatives? Well, we know that $\mathcal{L}\{f^{\prime\prime}(t)\} = s^2F(s) - sf(0) - f^{\prime}(0)$, which in our case would be $\mathcal{L}\{f^{\prime\prime}(t)\}=s^2F(s)-s$.

Therefore,

\[\mathcal{L}\{12t^2+4\} = \mathcal{L}\{f^{\prime\prime}(t)\} = s^2F(s)-s\]

We now compute the appropriate Laplace transforms and solve for $F(s)$:

\[s^2F(s) - s = \mathcal{L}\{12t^2+4\} = 12\cdot\frac{2!}{s^3}+4\cdot\frac{1}{s}=\frac{24}{s^3}+\frac{4}{s} \implies F(s) = \frac{24}{s^5} + \frac{4}{s^3}+\frac{1}{s}\]

Therefore, $\mathcal{L}\{(t^2+1)^2\} = \dfrac{24}{s^5}+\dfrac{4}{s^3}+\dfrac{1}{s}$.

I hope this makes sense!

- Jan 26, 2012

- 183

Question for the OP. Why would you want to compute this transform without expanding? I mean expanding would be the easiest approach.

BTW - Nice work Chris!

Following Chris's lead, you could keep going

\begin{align}

f(t) &= (t^2+1)^2,\;\;\; &f(0)&=1\\

f'(t) &= 4t(t^2+1)\;\;\; &f'(0)&=0\\

f''(t) &=4(t^2+1) + 8t^2,\;\;\;&f''(0)&=4\\

f'''(t) &= 8t + 16t,\;\;\;&f'''(0)&=0\\

f^{(4)}(t)&=24,\;\;\;&f^{(4)}(0)&=24\\

f^{(5)}(t)& = 0,

\end{align}

Then

$\mathcal{L}\{f^{(5)}(t)\} = s^5F(s) - f(0) s^4 - f'(0) s^3 - f''(0) s^2 - f'''(0) s - f^{(4)}(0) = 0.$

Substitute $f$ and its derivative at zero and solve for $F$ (same answer)

BTW - Nice work Chris!

Following Chris's lead, you could keep going

\begin{align}

f(t) &= (t^2+1)^2,\;\;\; &f(0)&=1\\

f'(t) &= 4t(t^2+1)\;\;\; &f'(0)&=0\\

f''(t) &=4(t^2+1) + 8t^2,\;\;\;&f''(0)&=4\\

f'''(t) &= 8t + 16t,\;\;\;&f'''(0)&=0\\

f^{(4)}(t)&=24,\;\;\;&f^{(4)}(0)&=24\\

f^{(5)}(t)& = 0,

\end{align}

Then

$\mathcal{L}\{f^{(5)}(t)\} = s^5F(s) - f(0) s^4 - f'(0) s^3 - f''(0) s^2 - f'''(0) s - f^{(4)}(0) = 0.$

Substitute $f$ and its derivative at zero and solve for $F$ (same answer)

Last edited:

- Thread starter
- #6