Suppose I want to make the substitution $\mathbf{k}\to-\mathbf{k}$ in the integral $$\int\mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(\mathbf{k},t)e^{-i\mathbf{k}\cdot\mathbf{r}}$$ where the domain of integration in $\mathbf{k}$-space is arbitrary. According to my course notes we should arrive at
$$\int\mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(\mathbf{k},t)e^{-i\mathbf{k}\cdot\mathbf{r}}\overset{\text{course notes}}{=}\int\mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(-\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{r}}$$
but I somehow think there should be an extra minus sign in this result, since I believe the substitution implies that $\mathrm{d}^3\mathbf{k}\to -\mathrm{d}^3\mathbf{k}$.
First let us write $\mathbf{k}'=-\mathbf{k}$, then $$\mathrm{d}^3\mathbf{k}=\mathrm{d}k_x\mathrm{d}k_y\mathrm{d}k_z=(-\mathrm{d}k_x')(-\mathrm{d}k_y')(-\mathrm{d}k_z')=-\mathrm{d}^3\mathbf{k}'$$ and $$\boldsymbol{\alpha}(\mathbf{k},t)e^{-i\mathbf{k}\cdot\mathbf{r}}=\boldsymbol{\alpha}(-\mathbf{k}',t)e^{i\mathbf{k}'\cdot\mathbf{r}}.$$ Hence \begin{align}\int\mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(\mathbf{k},t)e^{-i\mathbf{k}\cdot\mathbf{r}}&=-\int \mathrm{d}^3\mathbf{k}'\,\boldsymbol{\alpha}(-\mathbf{k}',t)e^{i\mathbf{k}'\cdot\mathbf{r}}\\ &=-\int \mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(-\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{r}}\end{align} where in the last line, I simply renamed the variable of integration. So,
What I think it should read: $$\int\mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(\mathbf{k},t)e^{-i\mathbf{k}\cdot\mathbf{r}}\overset{?}{=}-\int \mathrm{d}^3\mathbf{k}\,\boldsymbol{\alpha}(-\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{r}}.$$
I didn't mention the domain of integration and nor do the course notes, so I assume the arbitrariness of the domain of integration allows us to disregard the minus sign to arrive at the result stated in the course notes. I'm not quite sure, however, so I would really like some help.