Closed form for $\int_{-\infty}^\infty \frac{(x+1)\sin (3x)}{(x^2 + 6x + 10)^2} dx$

Question

Find the closed form for $$\mathcal{I} = \int_{-\infty}^\infty \frac{(x+1)\sin (3x)}{(x^2 + 6x + 10)^2} dx$$

My attempt

In order to find the closed form for this integral, what I first thought was using residue integration. First of all, notice two things

• The denominador cancels out when $$(x^2 + 6x + 10)^2 = 0 \Leftrightarrow x^2 + 6x + 10 = 0 \Leftrightarrow x = \frac{-6\pm \sqrt{6^2-4\cdot 10}}{2} = -3 \pm i$$ Non of these are in the real axis, and the only one on the upper semiplane is $-3+i$.
• From Euler's identity we notice that $e^{3x} = \cos(3x) + i \sin (3x)$. Therefore we can rewrite the integral as $$\mathcal{I} = \Im \left( \int_{-\infty}^\infty \frac{(x+1)e^{3ix}}{(x^2 + 6x + 10)^2} dx \right)$$

Now, from residue theory we know that, $$\int_{-\infty}^\infty \frac{(x+1)e^{3ix}}{(x^2 + 6x + 10)^2} dx = 2\pi i Res \left(\frac{(z+1)e^{3iz}}{(z^2 + 6z + 10)^2};z=-3+i\right)$$ The pole $-3+i$ has degree $2$, so $$ Res \left(\frac{(z+1)e^{3iz}}{(z^2 + 6z + 10)^2}; z=-3+i\right) = \lim_{z \to -3+i} \frac{d}{dz} \left\lbrace (z-(-3+i))^2\frac{(z+1)e^{3iz}}{(z^2 + 6z + 10)^2}\right\rbrace$$

This is where I get stuck since I can´t make it through the calculations. I think there might be other ways to tackle the problem or the limit, any help is truly appreciated :)

Answer 1

You should always do due diligence and check the semicircular contour vanishes; sometimes it does not. It does in this case, though.

The limit formula for a residue is correct but often a pain to handle. You've also incorrectly calculated the roots, it should be $-3+\color{red}{1}\cdot i$.

The pole is a double pole, and we can write the integrand as: $$(z-(-3+i))^{-2}\cdot\frac{z+1}{(z+3+i)^2}e^{3iz}$$We need to get the Taylor series of the righthand chunk at $-3+i$. Abbreviate $-3+i$ by $\zeta$. The numerator is $z+1=(z-\zeta)+(i-2)$. The denominator is: $$(z+3+i)^{-2}=-\frac{1}{4}-\frac{i}{4}(z-\zeta)+o(z-\zeta)$$The exponential is: $$e^{3iz}=e^{-3-9i}+3i\cdot e^{-3-9i}(z-\zeta)+o(z-\zeta)$$Multiplying them all together, we identify the first order term of the Taylor series as (see Cauchy product): $$\mathrm{Res}=-\frac{1}{4}\cdot e^{-3-9i}-\frac{i}{4}\cdot(i-2)\cdot e^{-3-9i}-\frac{1}{4}(i-2)\cdot3i\cdot e^{-3-9i}=\frac{e^{-3-9i}}{4}(3+8i)$$Multiplying this by $2\pi i$, we find: $$\frac{\pi\cdot e^{-3}}{2}(3i-8)e^{-9i}$$We can read off the imaginary part as: $$\int_{-\infty}^\infty\frac{(x+1)\cdot\sin(3x)}{(x^2+6x+10)^2}\,\mathrm{d}x=\frac{\pi}{2e^3}(3\cos(9)+8\sin(9))\approx0.0440732$$

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I have numerically confirmed this with the online version of Wolfram, just to double check I haven't erred; amusingly the calculator fails to evaluate this in a nice closed form.

Answer 2

Alternatively, integrate in real space with the substitution $y=x+3$

\begin{align} \mathcal{I} &= \int_{-\infty}^\infty \frac{(x+1)\sin (3x)}{(x^2 + 6x + 10)^2} dx\\ &=\cos9 \int_{-\infty}^\infty \frac{y\sin (3y)}{(y^2 + 1)^2} \overset{ibp}{dx} +2\sin9 \int_{-\infty}^\infty \frac{\cos (3y)}{(y^2 + 1)^2} dx \\ &=\frac{3\cos9}2 \int_{-\infty}^\infty \frac{\cos (3y)}{y^2 + 1} dx -\sin9 \ \frac{d}{da}\bigg(\int_{-\infty}^\infty \frac{\cos (3y)}{y^2 + a^2} dx \bigg)_{a=1} \end{align} Then, utilize $\int_{-\infty}^\infty \frac{\cos (3y)}{y^2 + a^2} dx= \frac\pi {a }e^{-3a}$ to obtain $$\mathcal{I} =\frac\pi{e^3}\left(\frac32\cos9+ 4\sin 9\right) $$