I came across the following step which I don't understand
$$\frac{1}{2\pi i}\int_{C}e^{v\left(\frac{1}{3}t^3-\eta t\right)}(t^2-\eta)g(t)\mathrm{d}t=-\frac{1}{2\pi i v}\int_{C}e^{v\left(\frac{1}{3}t^3-\eta t\right)}g'(t)\mathrm{d}t$$
And I know that the integral contains the Airy-function
$$\frac{1}{2\pi i}\int_{C}e^{v\left(\frac{1}{3}t^3-\eta t\right)}\mathrm{d}t=v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})$$
I tried integration by parts
\begin{align}\frac{1}{2\pi i}\int_{C}e^{v\left(\frac{1}{3}t^3-\eta t\right)}(t^2-\eta)g(t)\mathrm{d}t
&=v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\left[(t^2-\eta)g(t)\right]_C-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\int_{C}2t\ g(t)\mathrm{d}t\\
&-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\int_{C}(t^2-\eta)g'(t)\mathrm{d}t
\end{align}
Another integration by parts on the last two integrals (let $G(t):=\int g(t) \mathrm{d}t$)
\begin{align}
-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\int_{C}2t\ g(t)\mathrm{d}t&=-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})[2t\ G(t)]_C+v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})2\int_{C}\ G(t)\mathrm{d}t\\
\end{align}
\begin{align}
-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\int_{C}(t^2-\eta)g'(t)\mathrm{d}t
&=
-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})[(t^2-\eta)g(t)]_C+v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\int_{C}2t\ g(t)\mathrm{d}t
\\
\end{align}
Then we are left with
\begin{align}
-v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})[2t\ G(t)]+v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})2\int_{C}\ G(t)\mathrm{d}t+
v^{-1/3}\operatorname{Ai}(\eta\ v^{2/3})\int_{C}2t\ g(t)\mathrm{d}t
\\
\end{align}
If I make another integration by parts I end up with $0$, so clearly there is something wrong. Thank you for help
Best Answer
Let us integrate the derivative of function $$ f(t)= e^{v\left(\frac{1}{3}t^3-\eta t\right)} g(t) $$ over a contour $C$ such that $$\int_C f'(t) \,dt =0\,.$$ Such a contour can be a closed contour or a contour that goes to infinity in the directions in which $e^{v\left(\frac{1}{3}t^3-\eta t\right)}$ vanishes.
In this case from the product rule, we have that $$\int_C e^{v\left(\frac{1}{3}t^3-\eta t\right)} v (t^2-\eta) g(t) \,dt + \int_C e^{v\left(\frac{1}{3}t^3-\eta t\right)}g'(t)\,dt =0 $$ which is your result in a different form.