10.2: Integrals
- Page ID
- 6531
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Compute
\[I = \int_{-\infty}^{\infty} \dfrac{1}{(1 + x^2)^2}\ dx. \nonumber \]
Solution
Let
\[f(z) = 1/(1 + z^2)^2. \nonumber \]
It is clear that for \(z\) large
\[f(z) \approx 1/z^4. \nonumber \]
In particular, the hypothesis of Theorem 10.2.1 is satisfied. Using the contour shown below we have, by the residue theorem,
\[\int_{C_1 + C_R} f(z)\ dz = 2\pi i \sum \text{ residues of } f \text{ inside the contour.} \nonumber \]
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We examine each of the pieces in the above equation.
\(\int_{C_R} f(z)\ dz\): By Theorem 10.2.1(a),
\[\lim_{R \to \infty} \int_{C_R} f(z)\ dz = 0. \nonumber \]
\(\int_{C_1} f(z)\ dz\): Directly,we see that
\[\lim_{R \to \infty} \int_{C_1} f(z)\ dz = \lim_{R \to \infty} \int_{-R}^{R} f(x)\ dx = \int_{-\infty}^{\infty} f(x) \ dx = I. \nonumber \]
So letting \(R \to \infty\), Equation 10.3.4 becomes
\[I = \int_{-\infty}^{\infty} f(x) \ dx = 2\pi i \sum \text{ residues of } f \text{ inside the contour.} \nonumber \]
Finally, we compute the needed residues: \(f(z)\) has poles of order 2 at \(\pm i\). Only \(z = i\) is inside the contour, so we compute the residue there. Let
\[g(z) = (z - i)^2 f(z) = \dfrac{1}{(z + i)^2}. \nonumber \]
Then
\[\text{Res} (f, i) = g'(i) = -\dfrac{2}{(2i)^3} = \dfrac{1}{4i} \nonumber \]
So,
\[I = 2\pi i \text{Res} (f, i) = \dfrac{\pi}{2}. \nonumber \]
Compute
\[I = \int_{-\infty}^{\infty} \dfrac{1}{x^4 + 1} \ dx. \nonumber \]
Solution
Let \(f(z) = 1/(1 + z^4)\). We use the same contour as in the previous example (Figure \(\PageIndex{2}\).
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As in the previous example,
\[\lim_{R \to \infty} \int_{C_R} f(z)\ dz = 0 \nonumber \]
and
\[\lim_{R \to \infty} \int_{C_1} f(z)\ dz = \int_{-\infty}^{\infty} f(x) \ dx = I. \nonumber \]
So, by the residue theorem
\[I = \lim_{R \to \infty} \int_{C_1 + C_R} f(z) \ dz = 2\pi i \sum \text{ residues of } f \text{ inside the contour.} \nonumber \]
The poles of \(f\) are all simple and at
\[e^{i\pi /4}, e^{i3\pi /4}, e^{i5\pi /4}, e^{i7\pi /4}. \nonumber \]
Only \(e^{i\pi /4}\) and \(e^{i 3\pi /4}\) are inside the contour. We compute their residues as limits using L’Hospital’s rule. For \(z_1 = e^{i \pi /4}\):
\[\text{Res} (f, z_1) = \lim_{z \to z_1} (z - z_1) f(z) = \lim_{z \to z_1} \dfrac{z - z_1}{1 + z^4} = \lim_{z \to z_1} \dfrac{1}{4z^3} = \dfrac{1}{4e^{i 3\pi /4}} = \dfrac{e^{-i3\pi /4}}{4} \nonumber \]
and for \(z_2 = e^{i 3\pi /4}\):
\[\text{Res} (f, z_2) = \lim_{z \to z_2} (z - z_2) f(z) = \lim_{z \to z_2} \dfrac{z - z_2}{1 + z^4} = \lim_{z \to z_2} \dfrac{1}{4z^3} = \dfrac{1}{4e^{i 9\pi /4}} = \dfrac{e^{-i\pi /4}}{4} \nonumber \]
So,
\[I = 2\pi i (\text{Res} (f, z_1) + \text{Res} (f, z_2)) = 2\pi i (\dfrac{-1 - i}{4\sqrt{2}} + \dfrac{1 - i}{4\sqrt{2}}) = 2\pi i (-\dfrac{2i}{4\sqrt{2}}) = \pi \dfrac{\sqrt{2}}{2} \nonumber \]
Suppose \(b > 0\). Show
\[\int_{0}^{\infty} \dfrac{\cos (x)}{x^2 + b^2} \ dx = \dfrac{\pi e^{-b}}{2b}. \nonumber \]
Solution
The first thing to note is that the integrand is even, so
\[I = \dfrac{1}{2} \int_{-\infty}^{\infty} \dfrac{\cos (x)}{x^2 + b^2}. \nonumber \]
Also note that the square in the denominator tells us the integral is absolutely convergent.
We have to be careful because \(\cos (z)\) goes to infinity in either half-plane, so the hypotheses of Theorem 10.2.1 are not satisfied. The trick is to replace \(\cos (x)\) by \(e^{ix}\), so
\[\tilde{I} = \int_{-\infty}^{\infty} \dfrac{e^{ix}}{x^2 + b^2} \ dx, \ \ \ \ \text{with} \ \ \ \ I = \dfrac{1}{2} \text{Re} (\tilde{I}). \nonumber \]
Now let
\[f(z) = \dfrac{e^{iz}}{z^2 + b^2}. \nonumber \]
For \(z = x + iy\) with \(y > 0\) we have
\[|f(z)| = \dfrac{|e^{i(x + iy)}|}{|z^2 + b^2|} = \dfrac{e^{-y}}{|z^2 + b^2|}. \nonumber \]
Since \(e^{-y} < 1\), \(f(z)\) satisfies the hypotheses of Theorem 10.2.1 in the upper half-plane. Now we can use the same contour as in the previous examples (Figure \(\PageIndex{3}\).
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We have
\[\lim_{R \to \infty} \int_{C_R} f(z)\ dz = 0 \nonumber \]
and
\[\lim_{R \to \infty} \int_{C_1} f(z)\ dz = \int_{-\infty}^{\infty} f(x) \ dx = \tilde{I}. \nonumber \]
So, by the residue theorem
\[\tilde{I} = \lim_{R \to \infty} \int_{C_1 + C_R} f(z)\ dz = 2\pi i \sum \text{ residues of } f \text{ inside the contour.} \nonumber \]
The poles of \(f\) are at \(\pm bi\) and both are simple. Only \(bi\) is inside the contour. We compute the residue as a limit using L’Hospital’s rule
\[\text{Res} (f, bi) = \lim_{z \to bi} (z - bi) \dfrac{e^{iz}}{z^2 + b^2} = \dfrac{e^{-b}}{2bi}. \nonumber \]
So,
\[\tilde{I} = 2\pi i \text{Res} (f, bi) = \dfrac{\pi e^{-b}}{b}. \nonumber \]
Finally,
\[I = \dfrac{1}{2} \text{Re} (\tilde{I}) = \dfrac{\pi e^{-b}}{2b}, \nonumber \]
as claimed.
Be careful when replacing \(\cos (z)\) by \(e^{iz}\) that it is appropriate. A key point in the above example was that \(I = \dfrac{1}{2} \text{Re} (\tilde{I})\). This is needed to make the replacement useful.