Integral $\int_{-\infty}^{\infty}\frac{\sin(x-\frac{a}{x})}{(x+\frac{1}{x})}dx$ for this integral, I plan to solve it by using residues, but it seems when dealing with the integral over the upper semi-circle, it doesn't converge. Could you give some help, thank you very much!
$$\int_{-\infty}^{\infty}\frac{\sin(x-\frac{a}{x})}{(x+\frac{1}{x})}dx$$
 A: I am not going to use residues for this. This answer is going to abuse differentiation under the integral sign.
We start by noting that the integrand is even and thus,
$$I (a)=2 \int\limits_0^\infty\frac{\sin\big(x-\frac ax\big)}{x+\frac1x}\,\mathrm dx$$
Throughout the solution, I will be assuming that $a \geq 0$.
Differentiating w.r.t. $a$,
$$I'(a) = -2 \int\limits_0^\infty \frac{\cos\big(x-\frac ax\big)}{x^2+1}\,\mathrm dx $$
Again differentiating,
$$\begin{align}I' '(a) &= -2 \int\limits_0^\infty\frac{\sin\big(x-\frac ax\big)}{x(x^2+1)} \,\mathrm dx \\ &= -2 \int\limits_0^\infty\frac{\sin\big(x-\frac ax\big)}x-\frac{x\sin \big(x-\frac ax\big)}{x^2+1}\,\mathrm dx\end{align}$$
Using the substitution $x\mapsto \frac ax$, we can easily prove that $ \int\limits_0^\infty\frac{\sin(x-\frac ax)}x=0$. Thus,
$$I' '(a) = 2 \int\limits_0^\infty \frac{x\sin\big(x-\frac ax\big)}{x^2+1}\,\mathrm dx= I(a) $$
Now, we have a differential equation, cool!
$$\begin{align}I(a) &= I' '(a) \\ \implies I(a) &= k_1 e^a+k_2 e^{-a}  \end{align}$$
For evaluating the constants, we have the initial values $(I(0), I'(0))=\big(\frac\pi e, -\frac\pi e \big)$. Using this, we get $(k_1,k_2)=\big(0,\frac\pi e\big)$.
Thus, we conclude that
$$I(a) = \int\limits_{-\infty}^\infty\frac{\sin\big(x-\frac ax\big)}{x+\frac1x}\,\mathrm dx = \frac\pi {e^{a+1}} $$
