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Evaluate:

$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

Using only complex analysis.

$$I = \int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = (\frac{1}{2})\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

Consider a contour $C$ with a upper-axis semi-circle $B$ and the axis running from $-R \to R$

We will compute:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz$$

First:

$$z^2 + 1 = 0 \implies Z \in \{-i, i\}$$

Only, $z = i$ is in the semi circle region.

$$\text{Res}_{z=i} = \lim_{z \to i} (z-i)(f(i)) = \lim_{z \to i} \frac{\cos(z)}{z + i} = \frac{\cosh(1)}{2i}$$

Applying the residue theorem:

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = (2\pi i)\cdot \frac{\cosh(1)}{2i} = \pi\cdot\cosh(1)$$

$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$

But that is wrong, the answer for the full improper is:

$$\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = \frac{\pi}{e}$$

What am I doing wrong?

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  • $\begingroup$ You seem to be confused about the use of $e^{iz}$ in the given answers. Read: en.wikipedia.org/wiki/Euler%27s_formula $\endgroup$ – Gahawar Dec 22 '14 at 14:59
  • $\begingroup$ I am aware of eulers formula. But it states: $e^{iz} = \cos(z) + i\sin(z)$ it includes $\sin(z)$ as well. Which is why you cant just substitute $e^{iz} = \cos(z)$ which is wrong. It is true that Re[$e^{iz}$] = $\cos(z)$ but how can we implement that looking at the contour top half of plane. $\endgroup$ – Amad27 Dec 22 '14 at 15:04
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The cosine function does not vanish on the semicircle as $R \to \infty$; in fact, it does the opposite. You need to either 1) take the real part of $e^{i x}$ in the upper half plane, or 2) use $\cos{x} = (e^{i x}+e^{-i x})/2$ and use both the upper and lower half planes, respectively.

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  • $\begingroup$ so I chose the wrong contour? Can you please address how to choose a contour here: math.stackexchange.com/questions/1077554/…? It could help very much. $\endgroup$ – Amad27 Dec 22 '14 at 13:20
  • $\begingroup$ @Amad27: you didn't choose the wrong contour necessarily. You chose the wrong function to integrate over that contour. It seems to me, looking at your work, that you have omitted working out the piece that involves the integral over the semicircle. You really need to work it out to see what's going on. $\endgroup$ – Ron Gordon Dec 22 '14 at 13:21
  • $\begingroup$ But if we look at the function respectively. Should I have chosen a different contour? $\endgroup$ – Amad27 Dec 22 '14 at 13:22
  • $\begingroup$ @Amad27: please read my answer. If you choose option 1), then no, you don't have to. $\endgroup$ – Ron Gordon Dec 22 '14 at 13:23
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    $\begingroup$ @Amad27: actually, I'll pass. I'm a bit busy. If you're taking complex analysis and don't know what I am talking about, then you have some reading to do. $\endgroup$ – Ron Gordon Dec 22 '14 at 13:28
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we use $$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$

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  • $\begingroup$ Why $e^{iz}$ instead of $\cos(z)$?? $\endgroup$ – Amad27 Dec 22 '14 at 13:56
  • $\begingroup$ The integral on the semicircle of $e^{iz}$ goes to zero. The integral on the semicircle of $\cos z$ does not. $\endgroup$ – GEdgar Dec 22 '14 at 14:01
  • $\begingroup$ But the problem statement says cosine ??? $\endgroup$ – Amad27 Dec 22 '14 at 14:44
  • $\begingroup$ $\cos (z)$ is the real part of $e^{iz}$ $\endgroup$ – user50165 Dec 23 '14 at 23:00
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we use $$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$ then take real parts of the resulting integral .using the same contour $C$

enter image description here

\begin{array}{l} i \in C; - i \notin C \\ {\mathop{\rm Re}\nolimits} s\left( {f;i} \right) = \mathop {\lim }\limits_{z \to i} \left\{ {\left( {z - i} \right)\frac{{e^{iz} }}{{1 + z^2 }}} \right\} = \frac{{e^{ - 1} }}{{2i}} \\ \end{array} \begin{array}{l} \oint\limits_C {\frac{{e^{iz} }}{{1 + z^2 }}dz} = \int\limits_{ - R}^{ + R} {\frac{{e^{ix} dx}}{{1 + x^2 }}} + \int\limits_{S_R} {\frac{{e^{iz} dz}}{{1 + z^2 }}} = 2\pi i\frac{{e^{ - 1} }}{{2i}} = \frac{\pi }{e} \\ \Rightarrow \int\limits_{ - R}^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}} + i\int\limits_{ - R}^{ + R} {\frac{{\sin \left( x \right)dx}}{{1 + x^2 }} + } \int\limits_{S_R} {\frac{{e^{iz} dz}}{{1 + z^2 }}} = \pi e^{ - 1} \\ 2\int\limits_0^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}} = \pi e^{ - 1} ;\quad \left( {\int\limits_{S_R} {\frac{{e^{ix} dx}}{{1 + x^2 }} = 0} ;R \to + \infty \int\limits_{ - R}^{ + R} {\frac{{\sin \left( x \right)dx}}{{1 + x^2 }} = 0} } \right) \\ \mathop {\lim }\limits_{R \to + \infty } \int\limits_0^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}} = \frac{\pi }{2}e^{ - 1} \\ \end{array}

note: but if $y = {\mathop{\rm Im}\nolimits} \left( z \right)$ then $\cos \left( z \right) \approx \frac{{e^{\left| y \right|} }}{{\left| z \right|^2 }}$ for large $\left| z \right|$

we have the estimate $ \left| {\int\limits_{C_R } {\frac{{e^{iz} }}{{z^2 + 1}}dz} } \right| \le \int\limits_{C_R } {\frac{{e^{ - y} }}{{R^2 - 1}}\left| {dz} \right|} \le \frac{{\pi R}}{{R^2 - 1}} \to 0 $ as $R \to \infty $ where $ y = {\mathop{\rm Im}\nolimits} \left( z \right) > 0 $by the residue theorem $ \int\limits_{ - \infty }^{ + \infty } {\frac{{e^{ix} }}{{x^2 + 1}}dx = } 2\pi i\sum\limits_{{\mathop{\rm Im}\nolimits} a > 0} {{\mathop{\rm Re}\nolimits} s_a } \frac{{e^{iz} }}{{z^2 + 1}} = \frac{\pi }{e} $

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  • $\begingroup$ how come you aren't using $\cos(z)$ ?? $\endgroup$ – Amad27 Dec 22 '14 at 14:43

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