# Do we have $\int_0^t \frac{1}{2}tr(e^{\int_0^v f(u)Idu})dv=\int_0^te^{\int_0^vf(u)du}dv?$

If $$I$$ is a 2 by 2 identity matrix, do we have $$\int_0^t \frac{1}{2}tr(e^{\int_0^v f(u)Idu})dv=\int_0^te^{\int_0^vf(u)du}dv?$$ where $$tr$$ is the trace of the matrix.

How to understand this integral of a matrix? I know that $$\int_0^v f(u)Idu=\begin{bmatrix} \int_0^v f(u)du & 0 \\ 0 & \int_0^v f(u)du \end{bmatrix}$$ But what is $$tr(e^{\int_0^v f(u)Idu})=$$?

• $tr(e^{\int_0^v f(u)Idu})=\int_0^v f(u)du+\int_0^v f(u)du=2\int_0^v f(u)du.$
– Fred
Commented May 5, 2022 at 7:46
• @Fred Why the first equality holds? Is there exp of this integral? Commented May 5, 2022 at 19:18
• sorry, my comment is nonsense
– Fred
Commented May 5, 2022 at 20:13

Yes, and this works in any dimension $$d$$. Let $$g = g(v) = \int_0^v f$$. Then $$e^{\int_0^v f(u)\,I\,\mathrm d u} = e^{g\, I} = e^g\, I$$ This is because taking the exponential of a diagonal matrix is the same as taking the exponential on each of its diagonal coefficients. Therefore $$\mathrm{Tr}(e^{\int_0^v f(u)\,I\,\mathrm d u}) = d\, e^{-g}.$$