Two notes about notation and some philosophy of mathematics.
Notation: Mostly for the utility and elegance regarding the property of Duality, I prefer this consistent definition of the continuous Fourier Transform: $$ X(f) \triangleq \mathscr{F} \Big\{ x(t) \Big\} \triangleq \int\limits_{-\infty}^{+\infty} x(t) \, e^{-j 2 \pi f t} \ \mathrm{d}t $$ and inverse: $$ x(t) \triangleq \mathscr{F}^{-1} \Big\{ X(f) \Big\} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+j 2 \pi f t} \ \mathrm{d}f $$ Even with different signs on $j$, the elegant symmetry between the forward transform and inverse should be clear. $$ $$
While there is a quantitative difference between $-j$ and $+j$ (they are non-zero and they are negatives of each other), there is no qualitative difference between the two. They are both totally "imaginary", they have the same magnitude and, in reflection, they have the same angle. And most saliently, they are imaginary numbers that have equal claim to squaring to be $-1$. With any non-zero number, there are two square roots to $-1$ and both $-j$ and $+j$ have equal claim to being that $\sqrt{-1}$.
$$ $$It's like saying that clockwise and counterclockwise have equal claim to being "positive" rotation in mathematics. They are equivalent but opposite. Left and Right are similar. Whether the number line runs from left to right or opposite is just a matter of convention. Now this does not mean that the real numbers $+1$ and $-1$ are equivalent. One is the multiplicative identity and the other is not. But there is absolutely nothing essential you can say about $+j$ that is not also true for $-j$. You could go to every mathematics book and swap $+j$ and $-j$ everywhere and every theorem would be just as valid as it was before.
That said, it should be clear that there is no qualitative difference between the Fourier Transform and the Inverse Fourier Transform as expressed above. But that change in sign will mean a change in sign for either $f$ or $t$ (but not both, take your pick). Now if either $x(t)$ or $X(f)$ have even symmetry, then you need not worry about the change in sign with $t$ or $f$.
Like it's pretty easy to show that
$$ \mathscr{F} \Big\{ \operatorname{rect}(t) \Big\} = \operatorname{sinc}(f) $$
where
$$ \operatorname{rect}(u) \triangleq \begin{cases}
1 \qquad & \mathrm{for} \ |u| < \tfrac12 \\
\tfrac12 \qquad & \mathrm{for} \ |u| = \tfrac12 \\
0 \qquad & \mathrm{for} \ |u| > \tfrac12 \\
\end{cases} $$
and
$$ \operatorname{sinc}(u) \triangleq \begin{cases}
1 \qquad & \mathrm{for} \ u = 0 \\
\frac{\sin(\pi u)}{\pi u} \qquad & \mathrm{for} \ u \ne 0 \\
\end{cases} $$
So, accordingly,
$$ \mathscr{F}^{-1} \Big\{ \operatorname{sinc}(f) \Big\} = \operatorname{rect}(t) $$
Gee, I wonder what $ \mathscr{F} \Big\{ \operatorname{sinc}(t) \Big\} $ is?
As Matt pointed out, you did miss swapping the integral limits, but I would recommend changing your definitions to the above and your life with Duality will get a whole lot easier:
If $X(f) = \mathscr{F} \Big\{ x(t) \Big\}$, then $x(-f) = \mathscr{F} \Big\{ X(t) \Big\}$.
Your life with $X(0)$ and $x(0)$ and with Parseval's Theorem will get easier, too:
$$ X(0) = \int\limits_{-\infty}^{+\infty} x(t) \ \mathrm{d}t $$
$$ x(0) = \int\limits_{-\infty}^{+\infty} X(f) \ \mathrm{d}f $$
$$ \int\limits_{-\infty}^{+\infty} \Big| x(t) \Big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{+\infty} \Big| X(f) \Big|^2 \ \mathrm{d}f$$
NO nasty asymmetrical scaling factors to worry about!! (Just remember the $2\pi$ in the exponent.)
And you will be able to answer the question of how many functions of $t$ are exactly the same as their Fourier transform (with $f$ substituted for $t$). It's not just the Gaussian:
$$ x(t) = e^{-j \pi t^2} $$
which happens to have an identical Fourier Transform:
$$ X(f) = e^{-j \pi f^2} $$
Just everything in your continuous F.T. life will be easier if you use this notation.