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$$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$

This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?

Micah
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user85798
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  • You could probably try partial fraction decomposition on $\frac{x}{\left(x^2 + a\right)\left(x^2 + b\right)}$, but this is still going to require the integral of $\frac{\sin(x)}{x^2 + a^2}$ which isn't going to be easy (two applications of integration by parts may yield a result, but I doubt it). – Jared Apr 16 '14 at 00:51
  • WLOG if $a^2<b^2$ then you can write the numerator as $\displaystyle\frac{1}{b^2-a^2}(x^2+b^2-x^2-a^2)(xsin(x))$ and then break up accordingly. Note sure if this helps. – Joseph Zambrano Apr 16 '14 at 00:53
  • @JosephZambrano You should be able to assume $a^2 < b^2$ unless $a^2 = b^2$ (but then that's a special integral, right?). – Jared Apr 16 '14 at 00:55
  • It seems to me that you would have to write it in terms of the imaginary numbers which makes this very tedious (and not even necessarily possible) to integrate. – Jared Apr 16 '14 at 01:03
  • When Wolfram cannot take the integral, it's likely that a closed form integral (i.e. using "elementary" methods) is not possible. – Jared Apr 16 '14 at 01:08

3 Answers3

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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.

Lucian
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  • Is differentiation under the integral sign the best way to go about that integral? – user85798 Apr 16 '14 at 01:42
  • It's more a matter of personal taste or preference. It's neither simpler nor more difficult than other possible methods. – Lucian Apr 16 '14 at 02:22
  • I was curious about this problem, so I started with your method. I may have made a silly calculation error, but I get that your $I_n(t)$ evaluates to $(\pi / n) \cdot \cosh(tn)$ using the residue theorem with an upper-half plane semicircle contour. How did you evaluate it? – Matt R. Apr 19 '14 at 02:37
  • You seem to have left out a $-\sinh(tn)$. Then we have $\cosh x-\sinh x=e^{-x}$. – Lucian Apr 19 '14 at 03:10
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It may be good to try partial fractions. The result it:

$$ \int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)} dx = \frac{\pi}{a^2-b^2}\left(\sinh(a)-\cosh(a)+\cosh(b)-\sinh(b) \right) $$

Riemann1337
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2

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 $$ \int_0^\infty\frac{\cos(a t)}{t^2+1}dt=\frac{\pi}{2e^a}. $$

xpaul
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