I am currently trying to compute the following derivative and integral:
$$ P\psi_\theta = \frac{d}{d\theta}\int_{-k}^k tf_T(t)dt, $$ where $t=x-\theta$ and $X\sim N(\theta_0,\sigma^2)$. $f_T$ above is the density of $T$. It is part of a an expression, but the rest of the expression is $0$, so for everything to make sense, under $\theta_0$ it should evaluate to $$ |P\psi_{\theta_0}|=\int_{-k}^kf_T(t)dt. $$ You can find the source of this in Huber (1964), p. 79, in the numerator of $V$ for example (iii).
My attempt: We have that $dt=dx$, so we can simply substitute: $$ \frac{d}{d\theta}\int_{-k}^ktf_T(t)dt=\frac{d}{d\theta}\int_{-k}^k(x-\theta)f_X(x-\theta)dx=\frac{d}{d\theta}\left(\int_{-k}^k xf_X(x-\theta)dx-\theta\int_{-k}^kf_X(x-\theta)dx\right)\\ =-\int_{-k}^kxf'_X(x-\theta)dx-\int_{-k}^kf_X(x-\theta)dx+\theta\int_{-k}^kf'_X(x-\theta)dx\\ =-\int_{-k}^k (x-\theta)f'_X(x-\theta)dx-\int_{-k}^kf_X(x-\theta)dx. $$
So, the second term is what I want, thus meaning that the first should be $0$. But I don't think it is, since $$ -\int_{-k}^k (x-\theta)f'_X(x-\theta)dx=-\int_{-k}^k\frac{(x-\theta)^2}{\sigma^2}f_X(x-\theta)dx. $$ With an odd power, sure, but not when it's squared. Any ideas or hints? Thanks.
Actually, the change of variable $x=t+\theta$ yields $$\int_{-k}^k tf_T(t)dt=\int_{-k+\theta}^{k+\theta}(x-\theta)f_X(x)\mathrm dx$$ hence the derivative you are after is $$kf_X(k+\theta)+(-k)f_X(-k+\theta)+\int_{-k+\theta}^{k+\theta}(-1)f_X(x)\mathrm dx,$$ which is also $$k(f_T(k)-f_T(-k))-\int_{-k}^{k}f_T(t)\mathrm dt.$$ Thus, your suggestion holds if the first term is zero, that is, for example, if the density $f_T$ is even.