$\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}$ with using Residue theorem

Question

I solve this question with using Residue theorem. $$\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}$$

I think

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\begin{align}P\cdot V\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}&=2\pi i\left(\operatorname{Res}_{z=i}\left\{ \dfrac{1}{z^{2}+1}\right\}+\operatorname{Res}_{z=-i}\left\{ \dfrac{1}{z^{2}-1}\right\}\right)\\&=2\pi i\left( \lim _{z\rightarrow i}\left\{ \dfrac{1}{z+i}\right\} +\lim _{z\rightarrow -i}\left\{ \dfrac{1}{z-i}\right\} \right)\\&=2\pi i\left( \dfrac{1}{2i}+\left( -\dfrac{1}{2i}\right) \right) \\ &=0\end{align} But answer is $\pi$. This is answer. I don't know where I wrong. Please tell me.

Answer 1

You should only take into account those singularities with imaginary part greater than $0$. So, you will get $2\pi i\times\frac1{2i}$, which is indeed equal to $\pi$.

Answer 2

If I may expand on @JoséCarlosSantos's point, the idea is to use an "infinite semicircular contour", either counterclockwise on $\Im z\ge0$ or clockwise (thereby inducing a $-$ sign) on $\Im z\le0$. We thereby enclose the pole at $i$ or $-i$ respectively; either can be used, so "only one pole counts" isn't an arbitrary or special-pleasing statement. The pole at $\pm i$ can be used to evaluate the integral as$$\pm2\pi i\lim_{z\to\pm i}\frac{z\mp i}{z^2+1}=\pm2\pi i\lim_{z\to\pm i}\frac{1}{z\pm i}=\frac{\pm2\pi i}{\pm2i}=\pi.$$In fact, your original calculation can be salvaged: if we induce the sign change required of the contour enclosing the pole at $-i$, we're counting the integral twice (once with each contour), so the correct reuslt is $\tfrac122\pi i\left(\frac{1}{2i}\color{blue}{+}\frac{1}{2i}\right)=\pi$.

Answer 3

If you want to know why, in applying the residue theorem to calculate the principal value, you only need take the residue at the pole with positive imaginary part, You need to appl the residue theorem with care. You first choose a closed contour that traces the real axis, say from $-M$ to $+M$, and then circle back from $+M$ to $-M$ following a large semicircle or rectangle either above the real axis or below the real axis. In doing so it will enclose one or other of the two poles at $z = \pm i$ but not both. If I call this curve $\gamma$ you then have, from the residue theorem,
$$\int_{-M}^M \frac{1}{1+z^2} dz \pm \int_\gamma \frac{1}{1+z^2} dz = \pm 2\pi i \times R \tag{1} \label{E1}$$ where $R$ is the residue at whichever pole is enclosed, either at $z=i$ or $z=-i$. The sign ambiguity reflects the fact that in one choice, where $\gamma$ is taken above the real axis, the curve is traced counter clockwise so the plus sign is appropriate, and in the other the curve is traced clockwise so the minus sign becomes appropriate.

You then need to show that as $M \to \infty$ the integral along $\gamma$ has limit zero. This allows you to conclude that the integral along the axis, in the limit, is $\pm 2\pi i R$.

The trick is that the $\gamma$ integral often only converges to zero in one case and not the other. However, in this case you are lucky and the $z^2$ means you get a zero limit along the curve in either case and hence we should end up with either sign in \eqref{E1}, take the plus sign and $R$ to be the residue at $z = i$ or the minus sign and the residue at $z=-i$. This then gives the answer you wanted.