$$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$
This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?
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Sign up to join this community$$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$
This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?
Note $$\int_0^\infty e^{-xt}\cos(at)dt=\frac{x}{x^2+a^2}$$ and hence $$\int_0^\infty e^{-xt}(\cos(at)-\cos(bt))dt=-\frac1{a^2-b^2}\frac{x}{(x^2+a^2)(x^2+b^2)}.$$ Thus \begin{eqnarray} \int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx&=&2\int_{0}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx\\ &=&-\frac2{a^2-b^2}\int_{0}^\infty\left(\int_{0}^\infty e^{-xt}(\cos(at)-\cos(bt))dt\right)\sin xdx\\ &=&-\frac1{a^2-b^2}\int_{0}^\infty(\cos(at)-\cos(bt))\left(\int_{0}^\infty e^{-xt}\sin x dx\right)dt\\ &=&-\frac2{a^2-b^2}\int_{0}^\infty\frac{\cos(at)-\cos(bt)}{t^2+1}dt\\ &=&-\frac2{a^2-b^2}\left(\frac{\pi}{2e^a}-\frac{\pi}{2e^b}\right)\\ &=&-\frac1{a^2-b^2}\left(\frac{\pi}{e^a}-\frac{\pi}{e^b}\right). \end{eqnarray} Here we used the following result $$ \int_0^\infty\frac{\cos(a t)}{t^2+1}dt=\frac{\pi}{2e^a}. $$
Hint: First, prove that $I_n(t)=\displaystyle\int_{-\infty}^\infty\frac{\cos(tx)}{x^2+n^2}dx=\frac\pi n\cdot e^{-nt}$ . Then express your integral in terms of $I'_a(1)$ and $I'_b(1)$, since $\dfrac{d}{dt}\cos(tx)=-x\sin(tx)$, which, for $t=1$, becomes the numerator of our integrand.
It may be good to try partial fractions. The result it:
$$ \int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)} dx = \frac{\pi}{a^2-b^2}\left(\sinh(a)-\cosh(a)+\cosh(b)-\sinh(b) \right) $$