MHB Value of Sigma such that the integral converges

[Dustinsfl]

What value of sigma guarantees the integral converges?
\[
\int_0^{\infty}e^{-5t}e^{-(\sigma + i\omega)t}dt = \frac{1}{5 + \sigma + i\omega}
\]
I don't see a problem as long as \(\omega\neq 0\) and \(\sigma\neq -5\), but by the way the question is worded, it sounds like it is after something else since it doesn't mention the value of \(\omega\).

 

[chisigma]

What value of sigma guarantees the integral converges?
\[
\int_0^{\infty}e^{-5t}e^{-(\sigma + i\omega)t}dt = \frac{1}{5 + \sigma + i\omega}
\]
I don't see a problem as long as \(\omega\neq 0\) and \(\sigma\neq -5\), but by the way the question is worded, it sounds like it is after something else since it doesn't mention the value of \(\omega\).

Setting $s = \sigma + i\ \omega$ the integral becomes...

$\displaystyle \int_{0}^{\infty} e^{- 5\ t}\ e^{- s\ t}\ d t = \mathcal{L} \{e^{- 5\ t}\} = \frac{1}{s + 5}\ (1)$

... and the (1) converges for $\displaystyle \text{Re}\ s > - 5$...

Kind regards$\chi$ $\sigma$

 

[Dustinsfl]

Setting $s = \sigma + i\ \omega$ the integral becomes...

$\displaystyle \int_{0}^{\infty} e^{- 5\ t}\ e^{- s\ t}\ d t = \mathcal{L} \{e^{- 5\ t}\} = \frac{1}{s + 5}\ (1)$

... and the (1) converges for $\displaystyle \text{Re}\ s > - 5$...

Kind regards$\chi$ $\sigma$

What is wrong with \(\text{Re} \ s <-5\)?

 

[I like Serena]

What is wrong with \(\text{Re} \ s <-5\)?

An integral up to infinity of the form $\int_0^\infty e^{(x+iy)t} dt$ can only converge if $x<0$.
That is because the magnitude:
$$|e^{(x+iy)t}|=|e^{xt} e^{iyt}| = e^{xt}$$
will be of the form $e^{-t}$ if x is negative and $e^t$ if x is positive.
Only the first form will converge when integrated to infinity.

 

[Dustinsfl]

An integral up to infinity of the form $\int_0^\infty e^{(x+iy)t} dt$ can only converge if $x<0$.
That is because the magnitude:
$$|e^{(x+iy)t}|=|e^{xt} e^{iyt}| = e^{xt}$$
will be of the form $e^{-t}$ if x is negative and $e^t$ if x is positive.
Only the first form will converge when integrated to infinity.

The modulus of my problem is then
\[
\lvert e^{-(5+\sigma + i\omega)t}\rvert = e^{(5 + \sigma)t}
\]
since we lose the negative sign. Therefore, to have a negative exponential, wouldn't we need \(\sigma < - 5\)?

 

[I like Serena]

The modulus of my problem is then
\[
\lvert e^{-(5+\sigma + i\omega)t}\rvert = e^{(5 + \sigma)t}
\]
since we lose the negative sign. Therefore, to have a negative exponential, wouldn't we need \(\sigma < - 5\)?

We would not lose the negative sign.
$$|e^{-(5+\sigma + i\omega)t}| = |e^{-(5+\sigma)t} e^{i\omega t}| = |e^{-(5+\sigma)t} | \cdot |e^{i\omega t}| = |e^{-(5+\sigma)t}|$$
Since $e^{-(5+\sigma)t}$ is always a positive number, this is equal to:
$$e^{-(5+\sigma)t}$$

 

[Dustinsfl]

We would not lose the negative sign.
$$|e^{-(5+\sigma + i\omega)t}| = |e^{-(5+\sigma)t} e^{i\omega t}| = |e^{-(5+\sigma)t} | \cdot |e^{i\omega t}| = |e^{-(5+\sigma)}|$$
Since $e^{-(5+\sigma)t}$ is always a positive number, this is equal to:
$$e^{-(5+\sigma)t}$$

To take the modulus, we have \(\sqrt{(e^{-5t})^2} = e^{5t}\). We always lose the negative when doing this. Granted exponential is always greater than 0 but how does that preserve a negative which squared?

 

[I like Serena]

To take the modulus, we have \(\sqrt{(e^{-5t})^2} = e^{5t}\). We always lose the negative when doing this. Granted exponential is always greater than 0 but how does that preserve a negative which squared?

This is not true.
The square root cancels the square. The power $-5t$ remains intact.

Let's try it with t=1/5.
Then the argument is $e^{-1} \approx 0.368$.
If we square this number and then take the square root of the result, we will get the same number.
In particular we will not get the number $e^{1} \approx 2.71$