<s>[INST] Finding $\int _{-\infty }^{\infty }\frac{x^3\sin(x)}{x^4-b^4}\,\mathrm dx$ with real methods How can i prove with real methods that $\int_{-\infty}^\infty\frac{x^3\sin(x)}{x^4-b^4}\,\mathrm dx=\frac\pi2(e^{-b}+\cos(b))$? I was able to prove this using complex analysis but i dont know how to attack with without it.
As said in the comments we can try using Feynman's trick for this one:
$$I=\int _{-\infty }^{\infty }\frac{x^3\sin \left(x\right)}{x^4-b^4}\:dx$$
$$I\left(a\right)=\int _{-\infty }^{\infty }\frac{x^3\sin \left(ax\right)}{x^4-b^4}\:dx=\underbrace{\int _{-\infty }^{\infty }\frac{\sin \left(ax\right)}{x}\:dx}_{\pi }+b^4\int _{-\infty }^{\infty }\frac{\sin \left(ax\right)}{x\left(x^4-b^4\right)}\:dx$$
$$I'\left(a\right)=b^4\int _{-\infty }^{\infty }\frac{\cos \left(ax\right)}{x^4-b^4}\:dx$$
$$I''''\left(a\right)=b^4\underbrace{\int _{-\infty }^{\infty }\frac{x^3\sin \left(ax\right)}{x^4-b^4}\:dx}_{I\left(a\right)}$$
$$I''''\left(a\right)-b^4I\left(a\right)=0$$
And i think it boils down to this, i dont know if theres a more simpler or easier approach.
As xpaul showed, the general solution for that differential equation is $$I\left(a\right)=C_1e^{ab}+C_2e^{-ab}+C_3\cos \left(ab\right)+C_4\sin \left(ab\right)$$
But instead of letting $I\left(0\right)=0$ its better to use $I\left(0\right)=\pi $
thus having:
$$I\left(0\right)=C_1+C_2+C_3=\pi $$
$$I'\left(0\right)=C_1b-C_2b+C_4b=-\frac{\pi b}{2}$$
$$I''\left(0\right)=C_1b^2+C_2b^2-C_3b^2=0$$
$$I'''\left(0\right)=C_1b^3-C_2b^3-C_4b^3=-\frac{\pi b^3}{2}$$
with those mistakes fixed this eventually gives:
$$C_1=0,C_2=\frac{\pi }{2},C_3=\frac{\pi }{2},C_4=0$$
So,
$$I\left(a\right)=\frac{\pi }{2}\left(e^{-ab}+\cos \left(ab\right)\right)$$
$$I=\frac{\pi }{2}\left(e^{-b}+\cos \left(b\right)\right)$$
 [/INST] The characteristic equation of
$$ I''''(t)-b^4I(t)=0 \tag1$$
is
$$ r^4-b^4=0. \tag2$$
(2) has roots $\pm b,\pm bi$ and hence (1) has the general solution
$$ I(t)=C_1e^{bt}+C_2e^{-bt}+C_3\sin(bt)+C_4\cos(bt). $$
Noting 
$$I'\left(a\right)=b^4\int _{-\infty }^{\infty }\frac{\cos \left(ax\right)}{x^4-b^4}\:dx$$
one has $I(0)=I''(0)=0$ and
\begin{eqnarray}
I'\left(0\right)&=&b^4\int _{-\infty }^{\infty }\frac{1}{x^4-b^4}\:dx=\frac{b^2}{2}\bigg[\int _{-\infty }^{\infty }\frac1{x^2-b^2}dx-\int _{-\infty }^{\infty }\frac1{x^2+b^2}dx\bigg]=-\frac{b\pi}{2}\\
I'''(0)&=&-b^4\int _{-\infty }^{\infty }\frac{x^2}{x^4-b^4}\:dx=\frac{b^3\pi}{2}.
\end{eqnarray}
So
\begin{eqnarray}
C_1+C_2+C_4=0,\\
C_1b^2+C_2b^2-C_4b^2=0,\\
C_1b-C_2b+C_3b=-\frac{b\pi}{2},\\
C_1b^3-C_2b^3-C_3b^3=-\frac{b^3\pi}{2}
\end{eqnarray}
which implies that 
$$ C_1=C_3=0, C_2=C_4=\frac{\pi}{2}. $$
Thus
$$ I=I(1)=\frac{\pi}{2}(e^{-b}+\cos b).$$ 
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