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{% extends "base.html" %}
{% block title %}Area Under the Curve of a Complex Integral{% endblock %}
{% block style %}
<link href="/css/integral.css" rel="stylesheet">
{% endblock %}
{% block content %}
<h4>Area Under the Curve of a Complex Integral</h4>
<p>Scripts used to generate graphs: <a href="/dld/integral_scripts.zip">Link</a></p>
<h5>29/12/2018</h5>
<p>A definite integral can be represented on the xy-plane as the signed area
bounded by the curve of the function f(x), the x-axis, and the limits of
integration a and b. But it's not immediately clear how this definition applies
for complex valued functions.</p>
<p>Consider the following example:</p>
<div class="precontain"><div class="mathblock">
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<msubsup>
<mo>∫</mo>
<mn>0</mn>
<mn>1</mn>
</msubsup>
<msup>
<mfenced>
<mn>-1</mn>
</mfenced>
<mi>x</mi>
</msup>
<mi>dx</mi>
</mrow>
</math>
</div></div>
<p>If the function is graphed on the xy-plane, the real valued outputs are sparse.
Yet an elementary antiderivative exists and the definite integral is well defined.</p>
<figure>
<img src="/img/minus_one_exp_x_real_values_only.png"
alt="Real values only plot of minus one raised to the x power"
height="400"
width="520">
<figcaption>Figure 1 - Real values only</figcaption>
</figure>
<p>In order to plot a meaningful graph that can be used to potentially calculate
the integral as a signed area, some cues are taken from Philip Lloyd's work on
<a href="https://phantomgraphs.weebly.com/" target="_blank">Phantom Graphs</a>.
In that work, an additional z-axis is used to extend the x-axis into a complex
xz-plane, allowing complex inputs to be graphed. For the function considered
here, the z-axis is instead used to extend the y-axis into a complex yz-plane
to allow graphing of complex outputs instead.</p>
<p>Upon doing so, the following helical graph is obtained:</p>
<figure>
<img src="/img/minus_one_exp_x_full_plot.png"
alt="Complete three dimensional graph of all values of minux one raised to the x power"
height="400"
width="520">
<figcaption>Figure 2 - Full graph</figcaption>
</figure>
<p>The curve is continuous and spirals around the x-axis, intersecting with the
real xy-plane at the points plotted in the initial graph of Figure 1. However
it is still not clear how to represent the area under the curve.</p>
<p>Observing that complex numbers in cartesian form are composed of a real
part and an imaginary part, it is possible to decompose the function
into real and imaginary components. These are easy to obtain by rotating the
graph above to view the real and imaginary parts as flat planes.</p>
<table id="component">
<tr>
<td>
<figure>
<img src="/img/cos_pi_x.png"
alt="Graph of the real component of minus one raised to the power of x"
height="400"
width="520">
<figcaption>Figure 3 - Real component</figcaption>
</figure>
</td>
<td>
<figure>
<img src="/img/sin_pi_x.png"
alt="Graph of the imaginary component of minus one raised to the power of x"
height="400"
width="520">
<figcaption>Figure 4 - Imaginary component</figcaption>
</figure>
</td>
</tr>
</table>
<p>From this it can be seen that the function is a combination of a real valued
cosine and an imaginary valued sine. With the limits of integration under
consideration, the real values disappear and we are left with the following:</p>
<div class="precontain"><div class="mathblock">
<table>
<tr>
<td colspan="2">
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<msubsup>
<mo>∫</mo>
<mn>0</mn>
<mn>1</mn>
</msubsup>
<msup>
<mfenced>
<mn>-1</mn>
</mfenced>
<mi>x</mi>
</msup>
<mi>dx</mi>
</mrow>
</math>
</td>
</tr>
<tr>
<td>=</td>
<td>
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<mi>i</mi>
<msubsup>
<mo>∫</mo>
<mn>0</mn>
<mn>1</mn>
</msubsup>
<mi>sin</mi>
<mfenced>
<mrow>
<mi>π</mi>
<mo>⁢</mo>
<mi>x</mi>
</mrow>
</mfenced>
<mi>dx</mi>
</mrow>
</math>
</td>
</tr>
<tr>
<td>=</td>
<td>
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<mo>-</mo>
<mfrac>
<mi>i</mi>
<mi>π</mi>
</mfrac>
<msubsup>
<menclose notation="right">
<mrow>
<mi>cos</mi>
<mfenced>
<mrow>
<mi>π</mi>
<mo>⁢</mo>
<mi>x</mi>
</mrow>
</mfenced>
</mrow>
</menclose>
<mn>0</mn>
<mn>1</mn>
</msubsup>
</mrow>
</math>
</td>
</tr>
<tr>
<td>=</td>
<td>
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<mo>-</mo>
<mfrac>
<mi>i</mi>
<mi>π</mi>
</mfrac>
<mfenced>
<mrow>
<mn>-1</mn>
<mo>-</mo>
<mn>1</mn>
</mrow>
</mfenced>
</mrow>
</math>
</td>
</tr>
<tr>
<td>=</td>
<td>
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<mfrac>
<mrow>
<mn>2</mn>
<mi>i</mn>
</mrow>
<mi>π</mi>
</mfrac>
</mrow>
</math>
</td>
</tr>
</table>
</div></div>
<p>This agrees with the answer obtained by ordinary evaluation of the integral
without considering the graph, so the informal area under the curve definition
still works. Considering the area under the curve using polar coordinates also
works, but requires evaluating a less than pleasant infinite sum and so won't
be considered here.</p>
<p>The next interesting question is how this relates to the surface area of a
<a href="https://www.mathcurve.com/surfaces.gb/helicoiddroit/helicoiddroit.shtml" target="_blank">right helicoid</a>.</p>
{% endblock %}
|
