Does integration commute with substitution t=0?

[Happiness]

Let $g(x,t)=\int f(k,x,t)\,dk$

Under what conditions is the following true?

$g(x,0)=\int f(k,x,0)\,dk$

That is, we can get the value of $g(x,t)$ when $t=0$, by
(1) either substituting $t=0$ into $g(x,t)$ or
(2) by first substituting $t=0$ into $f(k,x,t)$ and then integrating wrt $k$.

Does it work for non-zero values of $t$?

EDIT: $g(x,t)=\int f(x,t)\,dx$ corrected to $g(x,t)=\int f(k,x,t)\,dk$.

 

[fresh_42]

Can you say something about the integration limits? As it stands, the integral is a function of $t$ and only of $t$. Thus $g(x,t)=\int f(x,t)dx = F(t)$ and $F(0) = g(x,0)$ as $t$ is treated as a constant in the integral.

 

[Happiness]

Can you say something about the integration limits? As it stands, the integral is a function of $t$ and only of $t$. Thus $g(x,t)=\int f(x,t)dx = F(t)$ and $F(0) = g(x,0)$ as $t$ is treated as a constant in the integral.

The integration is from $-\infty$ to $\infty$.

 

[fresh_42]

So the integration limits don't have a hidden dependency on $t$ and the above remains valid. The equation has to hold for all values of $t$ for which it is defined and thus especially for $t=0$.

 

[Happiness]

So the integration limits don't have a hidden dependency on $t$ and the above remains valid. The equation has to hold for all values of $t$ for which it is defined and thus especially for $t=0$.

Is the equation true because integration is a linear operator, i.e., $\int t\,f(x)\,dx = t\int f(x)\,dx$? If so, how can we proof it using linearity?

 

[fresh_42]

Is the equation true because integration is a linear operator, i.e. $\int t\,f(x)\,dx = t\int f(x)\,dx$? If so, how can we proof it using linearity?

No, it doesn't have anything to do with linearity and as $t$ is in the exponent, it even isn't linearly depending on $t$
It is much simpler: We have a function $\int f(x,t) dx$. But what if we wrote $f(x,t_0)=:h(x)$. As long as $t_0$ is constant, this is allowed. We haven't changed anything. The original equation can be thought of as: $\textrm{For all values } t=t_0 \textrm{ we have } g(x,t_0)=\int f(x,t_0)dx$ and therefore also for $t=t_0=0$. The integration has nothing to do with $t$ so $t$ is constant for the integration. One could even redefine $h(x)=f(x,t)$ and the integral would be the same, only that $t$ is now somewhere hidden in the definition of $h(x)$, say $h_t(x)$. Then $g(x,t)= \int h_t(x)dx$ again for all defined values $t$, e.g. for $t=0$.

I'm getting the feeling that the more I explain the more it might confuse you. At least it starts to confuse me. Maybe we should simply write $t=c$ and the entire equation looks a lot better: $g(x,c)=g_c(x)=\int f(x,c)dx = \int f_c(x)dx$.