I found this integral in a solution of heat equation: $$I(b)=\int_{0}^{\infty}e^{-z^2}\cos(bz)dz$$ Differentiating in b, $$I'(b)=\int_{0}^{\infty}(-ze^{-z^2})\sin(bz)dz$$
How is the differentiation done in the integral sign ? This looks too simple. But is it correct?
Hints: apply DCT. By MVT we have $\frac {e^{-z^{2}}-e^{-z^{2}}} {z-z'} =-2t e^{-t^{2}}$ for some $t$ between $z$ and $z'$. But $|-2t e^{-t^{2}}| \leq 2e^{-(z-1)^{2}} (z+1)$ if $|z-z'| <1$. Now show that $2e^{(z-1)^{2}} (z+1)$ is integrable. (Of course $\cos (bz)$ is bounded). Now DCT gives the justification.