# [SOLVED]continuous

#### dwsmith

##### Well-known member
How can I prove the below is continuous?

$$\int_{-\pi}^{\pi}te^{xt}\cos(yt)g(t)dt \quad\text{and}\quad -\int_{-\pi}^{\pi}te^{xt}\sin(yt)g(t)dt$$

define the fourier transform of g as
$$G(z) = \int_{-\pi}^{\pi}e^{zt}g(t)dt$$

We know t, e^{xt}, sine, and cosine are continuous which means their products are continuous but we don't know about g(t).

#### chisigma

##### Well-known member
How can I prove the below is continuous?

$$\int_{-\pi}^{\pi}te^{xt}\cos(yt)g(t)dt \quad\text{and}\quad -\int_{-\pi}^{\pi}te^{xt}\sin(yt)g(t)dt$$

define the fourier transform of g as
$$G(z) = \int_{-\pi}^{\pi}e^{zt}g(t)dt$$

We know t, e^{xt}, sine, and cosine are continuous which means their products are continuous but we don't know about g(t).
The definition of F.T. of g(*) You have given...

$\displaystyle \mathcal{F} \{g(t)\}= G(z)=\int_{- \pi}^{\pi} g(t)\ e^{z t}\ dt$ (1)

... implies that...

$\displaystyle G^{'}(z)= \int_{- \pi}^{\pi} t\ g(t)\ e^{z t}\ dt= \mathcal{F}\{t\ g(t)\}$ (2)

Now if $\displaystyle g(t)\ e^{z t}$ is integrable in $(-\pi\ ,\ \pi)$ the same is for $\displaystyle t\ g(t)\ e^{z t}$ so that G(z) is everywhere derivable and that means that G(z) is everywhere continuous...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
The definition of F.T. of g(*) You have given...

$\displaystyle \mathcal{F} \{g(t)\}= G(z)=\int_{- \pi}^{\pi} g(t)\ e^{z t}\ dt$ (1)

... implies that...

$\displaystyle G^{'}(z)= \int_{- \pi}^{\pi} t\ g(t)\ e^{z t}\ dt= \mathcal{F}\{t\ g(t)\}$ (2)

Now if $\displaystyle g(t)\ e^{z t}$ is integrable in $(-\pi\ ,\ \pi)$ the same is for $\displaystyle t\ g(t)\ e^{z t}$ so that G(z) is everywhere derivable and that means that G(z) is everywhere continuous...

Kind regards

$\chi$ $\sigma$
Can the derivative be slipped past the integral because of uniform continuous?

#### chisigma

##### Well-known member
Can the derivative be slipped past the integral because of uniform continuous?
The derivative of the function...

$\displaystyle g(z)= \int_{a}^{b} f(x,z)\ dx$ (1)

... exists if $f_{z} (x,z)$ exists and is continuous...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
Now if $\displaystyle g(t)\ e^{z t}$ is integrable in $(-\pi\ ,\ \pi)$ the same is for $\displaystyle t\ g(t)\ e^{z t}$ so that G(z) is everywhere derivable and that means that G(z) is everywhere continuous...
Don't we need the G' to be continuous because than the partials would be continuous?

#### chisigma

##### Well-known member
Don't we need the G' to be continuous because than the partials would be continuous?
We are interested only to the existence of G'(z), because if G'(z) exists then G(z) is continuous...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
We are interested only to the existence of G'(z), because if G'(z) exists then G(z) is continuous...

Kind regards

$\chi$ $\sigma$
I think you misunderstood the question. I want to show the first to integrals are continuous. I gave the Fourier Transform information because g(t) is defined by it.

#### chisigma

##### Well-known member
I think you misunderstood the question. I want to show the first to integrals are continuous. I gave the Fourier Transform information because g(t) is defined by it.
I'm sure to have misunderstood because continuous can be [if it is...] a function and not a definite integral...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
I'm sure to have misunderstood because continuous can be [if it is...] a function and not a definite integral...

Kind regards

$\chi$ $\sigma$
If the definite integrals are partial derivatives, then they can be continuous (or not) as well.