Question

Use Residue theorem to compute
Principle value
integral from ( - infinity to infinity )

[3x] / [ ( x^2 + 1 ) ^2 (x+2) ] dx

Answer 1

Answer on Question #50230 – Math - Complex Analysis

Use Residue theorem to compute Principle value integral from ( - infinity to infinity )

[3x] / [(x^2 + 1)^2 (x + 2)] dx

Solution

By definition, $P.V.\int_{-\infty}^{\infty}\frac{3zdz}{(z + 2)(z^2 + 1)^2} = \lim_{\varepsilon \to 0}\left(\int_{-\infty}^{-2 - \varepsilon}\frac{3zdz}{(z + 2)(z^2 + 1)^2} +\int_{-2 + \varepsilon}^{\infty}\frac{3zdz}{(z + 2)(z^2 + 1)^2}\right)$

We use the contour consisting of the segment from $-R$ to $-2 - \varepsilon$ along the real axis, followed by $-\gamma_{\varepsilon}$, the semicircle in the upper half plane, of radius $\varepsilon$. The contour $-\gamma_{\varepsilon}$ is clockwise, so that $\gamma_{\varepsilon}$ is counter clockwise. Then we move from $-2 + \varepsilon$ to $R$ along the real axis, followed by $\gamma_{R}$, the semicircle of radius $R$ in the upper half plane counter clockwise from $R$ to $-R$. Call the whole contour $\Gamma_{\varepsilon,R}$. The contour $\Gamma_{\varepsilon,R}$ is closed in the upper half plane and encircles the pole of order 2 at $z = i$.

By the Residue Theorem,

$$\begin{array}{l} \int_{\Gamma_{\varepsilon,R}} \frac{3 \mathrm{zd}z}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2} = 2 \pi \mathrm{i} \operatorname{res} \left[ \frac{3 \mathrm{z}}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2}, \mathrm{z} = \mathrm{i} \right] = 2 \pi \mathrm{i} \lim_{\mathrm{z} \to \mathrm{i}} \left( \frac{3 \mathrm{z}}{(\mathrm{z} + 2)(\mathrm{z} + \mathrm{i})^2} \right)' = \\ = 2 \pi \mathrm{i} \lim_{\mathrm{z} \to \mathrm{i}} \left( \frac{6 (-\mathrm{z}^2 - \mathrm{z} + \mathrm{i})}{(\mathrm{z} + 2)(\mathrm{z} + \mathrm{i})^3} \right) = 2 \pi \mathrm{i} \left( \frac{3}{25} + \frac{9}{100} \mathrm{i} \right). \end{array}$$

As for the top of the contour

$$\left| \int_{\Gamma_{R}} \frac{3 \mathrm{zd}z}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2} \right| \leq \frac{\pi \mathrm{R}^2}{(\mathrm{R}^2 + 1)^2 (\mathrm{R} + 2)} \rightarrow 0 \text{ as } R \rightarrow \infty.$$

As for small semicircle, we parameterize $\gamma_{\varepsilon}$ by $z(t) = -2 + \varepsilon t$, $0 \leq t \leq \pi$.

Then

$$\int_{\gamma_{\varepsilon}} \frac{3 \mathrm{zd}z}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2} = 3 \int_{0}^{\pi} \frac{\mathrm{i} \varepsilon \mathrm{e}^{\mathrm{i}t} \mathrm{d}t (-2 + \varepsilon \mathrm{e}^{\mathrm{i}t})}{\varepsilon \mathrm{e}^{\mathrm{i}t} \left( (-2 + \varepsilon \mathrm{e}^{\mathrm{i}t})^2 + 1 \right)^2} = 3 \mathrm{i} \int_{0}^{\pi} \frac{(-2 + \varepsilon \mathrm{e}^{\mathrm{i}t})}{(5 - 4 \varepsilon \mathrm{e}^{\mathrm{i}t} + \varepsilon^2 \mathrm{e}^{2\mathrm{i}t})^2} \mathrm{d}t \rightarrow \frac{-6 \pi \mathrm{i}}{25}, \text{ as } \varepsilon \rightarrow 0.$$

Combining all the parts, we have

$$2 \pi \mathrm{i} \left( \frac{3}{25} + \frac{9}{100} \mathrm{i} \right) = \lim_{\varepsilon \to 0} \lim_{\mathrm{R} \to \infty} \int_{\Gamma_{R,\varepsilon}} \frac{3 \mathrm{zd}z}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2} = \mathrm{P.V.} \int_{-\infty}^{\infty} \frac{3 \mathrm{zd}z}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2} - \left( -\frac{6\pi}{25} \mathrm{i} \right).$$

Then, $\mathrm{P.V.} \int_{-\infty}^{\infty} \frac{3 \mathrm{zd}z}{(\mathrm{z} + 2)(\mathrm{z}^2 + 1)^2} = 2 \pi \mathrm{i} \left( \frac{3}{25} + \frac{9}{100} \mathrm{i} \right) - \frac{6\pi}{25} \mathrm{i} = \frac{-9\pi}{50}$.

Answer: P. V. $\int_{-\infty}^{\infty} \frac{3 \cdot z \, dz}{(z + 2)(z^2 + 1)^2} = \frac{-9\pi}{50}$.

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