Let
$$I = \int_0^\infty\frac{x\sin 3x}{(x^2+16)^2}$$
Find $I$ using complex integration.
My try: We should consider $$f(z) = \frac{ze^{3iz}}{(z^2+16)^2}$$ Using WA, I found that $\operatorname{res}_{z = -4i} = \frac{-3e^{12}}{16}$ and $\operatorname{res}_{z = 4i} = \frac{3e^{-12}}{16}$. Therefore the value of $I$ should be $\pi (\frac{-3e^{12}}{16} + \frac{3e^{-12}}{16})$ according to the residue theorem but WA gives $I = \frac{3\pi}{16e^{12}}$. What am I missing here? Also I wanted to know other methods for evaluating this integral like differentiation under the integral sign. I tried $\sin \lambda x$ and then differentiate with respect to $\lambda$ but it got messy.
Here is a method that mix Residue Theorem and Differentiation Under the Integral Sign. First, let's consider the following function: $$F(\alpha,\beta)=\int_0^\infty\frac{\cos (\alpha z)}{z^2+\beta^2}dz=\frac{1}{2}\mathfrak{R}\left(\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z^2+\beta^2}dz\right)=\frac{1}{2}\mathfrak{R}\left(2\pi i \frac{e^{-\alpha\beta}}{2\beta i} \right)$$
$$F(\alpha,\beta)=\int_0^\infty\frac{\cos (\alpha z)}{z^2+\beta^2}dz=\frac{\pi}{2\beta }e^{-\alpha\beta}$$
$$\frac{\partial^2F(\alpha,\beta)}{\partial\alpha\partial\beta}=2\beta\int_0^\infty\frac{z\sin (\alpha z)}{(z^2+\beta^2)^2}dz=\frac{\pi\alpha}{2 }e^{-\alpha\beta}$$
Another approach using only Differentiation Under the Integral Sign: $$F(\alpha,\beta)=\int_0^\infty\frac{\sin (\alpha z)}{z(z^2+\beta^2)}dz=\frac{1}{\beta^2}\int_0^\infty \underbrace{\frac{\sin (\alpha z)}{z}dz}_{\frac{\pi}{2}}+\frac{1}{\beta^2}\int_0^\infty \underbrace{-\frac{z\sin (\alpha z)}{z^2+\beta^2}dz}_{\frac{\partial^2 F}{\partial \alpha^2}}$$
$$\frac{\partial^2 F(\alpha,\beta)}{\partial \alpha^2}-\beta^2 F(\alpha,\beta)=-\frac{\pi}{2}$$ $$F_c(\alpha,\beta)=c_1e^{\alpha\beta}+c_2e^{-\alpha\beta}$$
$$F_p=(\alpha,\beta)=-e^{\alpha\beta}\int\frac{-\frac{\pi}{2}e^{-\alpha\beta}}{W(e^{\alpha\beta},e^{-\alpha\beta})}d\alpha+e^{-\alpha\beta}\int\frac{-\frac{\pi}{2}e^{\alpha\beta}}{W(e^{\alpha\beta},e^{-\alpha\beta})}d\alpha$$
$$F_p=(\alpha,\beta)=-\frac{\pi e^{\alpha\beta}}{4\beta}\int e^{-\alpha\beta}d\alpha+\frac{\pi e^{-\alpha\beta}}{4\beta}\int e^{\alpha\beta}d\alpha=\frac{\pi}{2 \beta^2}$$
Therefore: $$F(\alpha,\beta)=\int_0^\infty\frac{\sin (\alpha z)}{z(z^2+\beta^2)}dz=c_1e^{\alpha\beta}+c_2e^{-\alpha\beta}+\frac{\pi}{2 \beta^2}$$
$$F(0,\beta)=0=c_1+c_2+\frac{\pi}{2 \beta^2};\frac{\partial F(0,\beta)}{\partial\alpha}=\frac{\pi}{2\beta}=\beta(c_1-c_2)$$
Thus: $$F(\alpha,\beta)=\int_0^\infty\frac{\sin (\alpha z)}{z(z^2+\beta^2)}dz=-\frac{\pi}{2 \beta^2}e^{-\alpha\beta}+\frac{\pi}{2 \beta^2}$$
$$\frac{\partial^3F(\alpha,\beta)}{\partial\alpha^2\partial\beta}=2\beta\int_0^\infty\frac{z\sin (\alpha z)}{(z^2+\beta^2)^2}dz=\frac{\pi\alpha}{2 }e^{-\alpha\beta}$$