Show an integral of continuous $\gamma:\mathbb{R}\to GL(n,\mathbb{R})$ is invertible. 
Let
  $$\gamma:\mathbb{R}\to GL(n,\mathbb{R})$$
  be a continuous map such that
  $$\gamma(s+t)=\gamma(s)\gamma(t),\quad\gamma(0)=I,$$
  for all $s,t\in\mathbb{R}$, and let $\psi:\mathbb{R}\to[0,\infty)$ be a smooth function with compact support such that
  $$\int_{-\infty}^\infty\psi(t)dt=1.$$
  How to show that
  $$\int_{-\infty}^\infty\psi(t)\gamma(t)dt\in GL(n,\mathbb{R})?$$
  EDIT: Thanks to Robert Israel, it is in fact not true...

I was thinking that if this is not true then there exists $v\in\mathbb{R}^n$, $v\neq 0$ such that $\int_{-\infty}^\infty \psi(t)\gamma(t)vdt=0$. But $\gamma(t)v\neq 0$ for all $t$, so we have
$$\int_{-\infty}^\infty \psi(t)a_1(t)dt=0,\quad\ldots,\quad\int_{-\infty}^\infty\psi(t)a_n(t)dt=0,$$
for some continuous functions $a_i:\mathbb{R}\to\mathbb{R}$ such that for each $t_0\in\mathbb{R}$ at least one of them is non-zero. So we can take $t_0$ such that $\psi(t_0)>0$, but then it does not necessarily lead to a contradiction since the integrals are $0$ only on the whole domain $(-\infty,\infty)$.
 A: (EDITED)
Not true.  Take $n=2$ and $\gamma(t) = \pmatrix{\cos(t) & \sin(t)\cr -\sin(t) & \cos(t)\cr}$.  Take $\psi$ such that $\int_{-\infty}^\infty \psi(t)\cos(t)\; dt = \int_{-\infty}^\infty \psi(t) \sin(t)\; dt = 0$ (i.e. the Fourier transform of $\psi$ is $0$ at $1$ and $-1$).  The indicator function of the interval $[0,2\pi]$ would work, except that it's not smooth.  An appropriate smoothing should work: $\psi(t) = \int_{0}^{2\pi} \phi(t+s)\; ds$ where $\phi$ is smooth, nonnegative and compactly supported.
A: Robert israel's answer explain the problem. I'll just show how to get a smooth function $\psi$, by using partitions of unity.
Topologically, $SO_2 (\mathbb{R}) \sim \mathbb{S}_1$. The circle is smooth, andthere is an open cover given by $U_1 := \mathbb{S}_1 \setminus \{(1, 0)\}$ and $U_2 := \mathbb{S}_1 \setminus \{(-1, 0)\}$. Hence, we can find two smooth functions $\varphi_1, \varphi_2$ in $\mathcal{C}^\infty (\mathbb{S}_1, [0,1])$ such that:


*

*$\varphi_1 + \varphi_2 = 1$,

*The support of $\varphi_i$ is in $U_i$ for $i \in \{1,2\}$.
Now, Let $\gamma_1$ be the restriction of $\gamma$ to $I_1 := (-\pi, \pi)$, and $\gamma_2$ the restriction of $\gamma$ to $I_2 := (0, 2 \pi)$. Then each $\gamma_i$ is a diffeomorphism from $I_i$ to $U_i$. Let $2 \pi \psi := 1_{I_1} \cdot \varphi_1 \circ \gamma_1 + 1_{I_2} \cdot \varphi_2 \circ \gamma_2$. Then $\psi$ is smooth, takes its values in $[0,1]$, has compact support, and $\int_\mathbb{R} \psi = 1$. By construction, $\int_\mathbb{R} \psi \cdot \gamma$ is the null matrix.
