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I wish to evaluate the integral $\int_{-\infty}^\infty \frac{1}{(4+x^2)^2}dx $. This is an complex analysis exercise so expecting that some sort of complex analysis technique is required, my first instinct was to split the integrand into partial fractions, making use of the fact that $(4+x^2)^2=(x+2i)^2(x-2i)^2$ but the Cauchy Integral Formula does not seem to work here so I got stuck. Then I went back to using substitution, say x = tan u but that didn't really work out as well, though I am not sure if I missed any technical details.

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  • $\begingroup$ If you let $x=2\tan u$ then you will have to calculate an integral of $\cos^2u$. Just commenting, since you seem to have thought about such substitutions as well. $\endgroup$ – mickep Mar 30 '17 at 8:40
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One way to simplify the approach is to use differentiation under the integral. Then, note that

$$-\frac1{2a}\frac{d}{da}\int_{-\infty}^\infty\frac{1}{a^2+x^2}\,dx=\int_{-\infty}^\infty\frac{1}{(a^2+x^2)^2}\,dx\tag 1$$

The integral on the left-hand side of $(1)$ can be evaluate a number of ways. Inasmuch as the OP seeks to use contour integration, we proceed as follows.

Let $C$ be the contour comprised of the real line segment from $-R$ to $R$ and the upper-half-plane semi-circle centered at $0$ with radius $R$. Then, we have from the residue theorem for $R>a>0$

$$\begin{align} \oint_C \frac{1}{z^2+a^2}\,dz&=\int_{-R}^R \frac{1}{x^2+a^2}\,dx+\int_0^\pi \frac{1}{(Re^{i\phi})^2+a^2}\,iRe^{Re^{i\phi}}\,d\phi \tag 2\\\\ &=2\pi i \text{Res}\left(\frac{1}{z^2+a^2},z=ia\right)\\\\ &=\frac{\pi }{a} \end{align}$$

The second integral on the right-hand side of $(2)$ vanishes as $R\to \infty$. Hence, we find that

$$\int_{-\infty}^\infty \frac{1}{x^2+a^2}\,dx=\frac\pi a \tag 3$$

Taking the derivative of $(3)$ with respect to $a$, multiplying by $-1/(2a)$ and setting $a=2$ yields the coveted result

$$\int_{-\infty}^\infty\frac{1}{(4+x^2)^2}\,dx=\frac\pi{16}$$

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  • $\begingroup$ I noted that f(z) is an even function so I tried to calculate Res(f(z); 2i) directly by forming the partial fractions, thinking that multiplying the residue with 2πi would give me the final answer but it didn't work. Do you know why? $\endgroup$ – Homaniac Apr 2 '17 at 21:19
  • $\begingroup$ @homaniac That should have worked. How did you determine the residue? The pole in the upper-half (and the one is the lower half-plane) is of second order. $\endgroup$ – Mark Viola Apr 3 '17 at 1:13
  • $\begingroup$ Thanks for the feedback. Something went wrong with the way I split the partial fractions, my usual way of doing it got me stuck. Has that got something to do with it being of second order? $\endgroup$ – Homaniac Apr 3 '17 at 5:57
  • $\begingroup$ @homaniac If you would, show your decomposition into partial fractions. $\endgroup$ – Mark Viola Apr 3 '17 at 13:39
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This is the standard approach to a problem like yours: Integrate over a semicircle with diameter along the real axis and center at $0$, using the residue theorem (or Cauchy's integral formula of you can). Let the radius go towards $\infty$, and argue that the contribution to the integral from the circle arc goes to $0$.

No partial fractions needed, although the factorisation is handy when locating and classifying poles.

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An alternative solution through real-analytic tools. For any $\alpha>0$ we have $$ I(\alpha) = \int_{-\infty}^{+\infty}\frac{1}{\alpha^2+x^2}\,dx =\frac{\pi}{\alpha} \tag{1} $$ and by differentiating both sides with respect to $\alpha$ we get: $$ I'(\alpha) = -\int_{-\infty}^{+\infty}\frac{2\alpha}{(\alpha^2+x^2)^2}\,dx = -\frac{\pi}{\alpha^2}\tag{2} $$ hence by evaluating at $\alpha=2$ we get: $$ \int_{-\infty}^{+\infty}\frac{dx}{(4+x^2)^2} = \left.\frac{\pi}{2\alpha^3}\right|_{\alpha=2}=\color{red}{\frac{\pi}{16}}.\tag{3}$$

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  • $\begingroup$ I'm shocked. Jack D'Aurizio takes the trivial way. ;-)). I also used Differentiation under the integral, but evaluated the trivial integral using contour integration. $\endgroup$ – Mark Viola Mar 30 '17 at 20:13
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Another alternative is to use Glaisher's theorem. If $f(x)$ is an even function with series expansion of the form:

$$f(x)= \sum_{k=0}^{\infty}(-1)^kc_k x^{2k}$$

and the integral over the real line converges, then we have:

$$\int_{-\infty}^{\infty}f(x) dx = \pi c_{-\frac{1}{2}}$$

where $c_{-\frac{1}{2}}$ is defined from an appropriate analytic expression of the series expansion coefficients, where factorials are supposed to be replaced by gamma functions. This can be made more rigorous, it's a special case of Ramanujan's master theorem. The advantage of this method is that you essentially have the same benefit as when doing complex analysis where the integral is derived from an appropriate series expansion coefficient, but you now don't have to bother about setting up appropriate contours. In fact, the theorem remains valid if contour integration won't work.

So, all we have to do is find an analytic expression for $c_k$. This is a simple matter of differentiating the geometric series. If we put $x^2 = y$, then we have:

$$\frac{1}{4+y} = \sum_{k=0}^{\infty}(-1)^k \frac{y^k}{4^{k+1}}$$

Taking minus the derivative w.r.t.$y$ yields:

$$\frac{1}{(4+y)^2} = -\sum_{k=1}^{\infty}(-1)^k k \frac{y^{k-1}}{4^{k+1}} =\sum_{k=0}^{\infty}(-1)^k (k+1) \frac{y^k}{4^{k+2}} $$

So, we see that $c_k = \frac{k+1}{4^{k+2}}$ and we're now supposed to analytically continue this to real $k$ in the standard way allowing us to insert $k = -\frac{1}{2}$ in here, but in this case that's trivial. This yields $c_{-\frac{1}{2}}=\frac{1}{16}$, the integral is thus $\frac{\pi}{16}$.

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  • $\begingroup$ I noted that f(z) is an even function so I tried to calculate Res(f(z); 2i) directly by forming the partial fractions, thinking that multiplying the residue with 2πi would give me the final answer but it didn't work. Do you know why? $\endgroup$ – Homaniac Apr 2 '17 at 21:20
  • $\begingroup$ @Homaniac That residue is $-\frac{i}{32}$, so you may have a mistake computing the partial fraction expansion. But note that you can compute the residue directly and in fact a fast way to get to a partial fraction expansion is by computing residues, or at least the Laurent expansions needed to compute residues. $\endgroup$ – Count Iblis Apr 2 '17 at 21:35
  • $\begingroup$ I have 1/(x-2i)^2(x+2i)^2 = 1/8ix/(x-2i)^2 - 1/8ix/(x+2i)^2 and I got stuck here :( $\endgroup$ – Homaniac Apr 2 '17 at 21:57
  • $\begingroup$ @Homaniac Suppose $R(z)$ is a rational function that tends to zero at infinity, with poles at $z = z_j$. Then consider the singular parts of the expansions around each $z_j$, let's call these truncated expansions $S_j(z)$. Then the partial fraction expansion of $R(z)$ is $\sum_j S_j(z)$. You can see this by considering the function $f(z) = R(z) - \sum_j S_j(z)$. Then $f(z)$ is a rational function that tends to zero at infinity, but which has only removable singularities. So, $f(z)$ is a polynomial, but since it tends to zero at infinity, it is in fact identical to zero. So, let's try this. $\endgroup$ – Count Iblis Apr 2 '17 at 22:40
  • $\begingroup$ We can simplify things in this case by considering $R(z) = \frac{1}{4 + z^2}$, and then square each expansion, we then need to consider the constant term as well, as that generates a singular term upon squaring. If we put $z = 2 i + t$ and expand in powers of $t$, we get: $$\frac{1}{4it + t^2} = \frac{1}{4it}\left(1+\frac{it}{4}+\mathcal{O}(t^2)\right)$$ Squaring this yields a singular part of $R^2(z)$ of $$-\frac{1}{16 t^2}-\frac{i}{32 t}$$ $\endgroup$ – Count Iblis Apr 2 '17 at 22:47

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