Integration using Euler's formula: Difference between revisions

Revision as of 15:47, 30 August 2008

Latest revision as of 04:50, 12 July 2025

(99 intermediate revisions by 62 users not shown)

Line 1:

{{short description|Use of complex numbers to evaluate integrals}}

#REDIRECT[[Methods of contour integration]]

{{more citations needed|date=July 2019}}

{{calculus|expanded=integral}}

In [[integral calculus]], [[Euler's formula]] for [[complex number]]s may be used to evaluate [[integral]]s involving [[trigonometric functions]]. Using Euler's formula, any trigonometric function may be written in terms of complex exponential functions, namely <math>e^{ix}</math> and <math>e^{-ix}</math> and then integrated. This technique is often simpler and faster than using [[trigonometric identities]] or [[integration by parts]], and is sufficiently powerful to integrate any [[rational fraction|rational expression]] involving trigonometric functions.<ref>{{Cite journal|last=Kilburn|first=Korey|title=Applying Euler's Formula to Integrate|journal=American Review of Mathematics and Statistics|date=2019 |publisher=American Research Institute for Policy Development|volume=7|pages=1–2|doi=10.15640/arms.v7n2a1|doi-broken-date=12 July 2025 |issn=2374-2348|eissn=2374-2356|doi-access=free|url=https://arms.thebrpi.org/vol-7-no-2-december-2019-abstract-1-arms |hdl=2158/1183208|hdl-access=free}}</ref>

==Euler's formula==

Euler's formula states that<ref>{{Cite web|last=Weisstein|first=Eric W.|title=Euler Formula|url=https://mathworld.wolfram.com/EulerFormula.html|access-date=2021-03-17|website=mathworld.wolfram.com|language=en}}</ref>

:<math>e^{ix} = \cos x + i\,\sin x.</math>

Substituting <math>-x</math> for <math>x</math> gives the equation

:<math>e^{-ix} = \cos x - i\,\sin x</math>

because cosine is an even function and sine is odd. These two equations can be solved for the sine and cosine to give

:<math>\cos x = \frac{e^{ix} + e^{-ix}}{2}\quad\text{and}\quad\sin x = \frac{e^{ix}-e^{-ix}}{2i}.</math>

==Examples==

=== First example ===

Consider the integral

:<math>\int \cos^2 x \, dx .</math>

The standard approach to this integral is to use a [[half-angle formula]] to simplify the integrand. We can use Euler's identity instead:

:<math>\begin{align}

\int \cos^2 x \, dx \,&=\, \int \left(\frac{e^{ix}+e^{-ix}}{2}\right)^2 dx \\[6pt]

&=\, \frac14\int \left( e^{2ix} + 2 +e^{-2ix} \right) dx

\end{align}</math>

At this point, it would be possible to change back to real numbers using the formula {{math|''e''^2''ix'' + ''e''^−2''ix'' {{=}} 2 cos 2''x''}}. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:

:<math>\begin{align}

\frac14\int \left( e^{2ix} + 2 + e^{-2ix} \right) dx

&= \frac14\left(\frac{e^{2ix}}{2i} + 2x - \frac{e^{-2ix}}{2i}\right)+C \\[6pt]

&= \frac14\left(2x + \sin 2x\right) +C.

\end{align}</math>

===Second example===

Consider the integral

:<math>\int \sin^2 x \cos 4x \, dx.</math>

This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless:

:<math>\begin{align}

\int \sin^2 x \cos 4x \, dx

&= \int \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2\left(\frac{e^{4ix}+e^{-4ix}}{2}\right) dx \\[6pt]

&= -\frac18\int \left(e^{2ix} - 2 + e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right) dx \\[6pt]

&= -\frac18\int \left(e^{6ix} - 2e^{4ix} + e^{2ix} + e^{-2ix} - 2e^{-4ix} + e^{-6ix}\right) dx.

\end{align}</math>

At this point we can either integrate directly, or we can first change the integrand to {{math|2 cos 6''x'' − 4 cos 4''x'' + 2 cos 2''x''}} and continue from there.

Either method gives

:<math>\int \sin^2 x \cos 4x \, dx = -\frac{1}{24} \sin 6x + \frac18\sin 4x - \frac18 \sin 2x + C.</math>

==Using real parts==

In addition to Euler's identity, it can be helpful to make judicious use of the [[real part]]s of complex expressions. For example, consider the integral

:<math>\int e^x \cos x \, dx.</math>

Since {{math|cos ''x''}} is the real part of {{math|''e''^''ix''}}, we know that

:<math>\int e^x \cos x \, dx = \operatorname{Re}\int e^x e^{ix}\, dx.</math>

The integral on the right is easy to evaluate:

:<math>\int e^x e^{ix} \, dx = \int e^{(1+i)x}\,dx = \frac{e^{(1+i)x}}{1+i} + C.</math>

Thus:

:<math>\begin{align}

\int e^x \cos x \, dx &= \operatorname{Re}\left(\frac{e^{(1+i)x}}{1+i}\right) + C \\[6pt]

&= e^x\operatorname{Re}\left(\frac{e^{ix}}{1+i}\right) +C \\[6pt]

&= e^x\operatorname{Re}\left(\frac{e^{ix}(1-i)}{2}\right) +C \\[6pt]

&= e^x \frac{\cos x + \sin x}{2} +C.

\end{align}</math>

==Fractions==

In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral

:<math>\int \frac{1+\cos^2 x}{\cos x + \cos 3x} \, dx.</math>

Using Euler's identity, this integral becomes

:<math>\frac12 \int \frac{6 + e^{2ix} + e^{-2ix} }{e^{ix} + e^{-ix} + e^{3ix} + e^{-3ix}} \, dx.</math>

If we now make the [[integration by substitution|substitution]] <math>u = e^{ix}</math>, the result is the integral of a [[rational function]]:

:<math>-\frac{i}{2}\int \frac{1+6u^2 + u^4}{1 + u^2 + u^4 + u^6}\,du.</math>

One may proceed using [[partial fraction decomposition]].

==See also==

{{Portal|Mathematics}}

* [[Trigonometric substitution]]

* [[Weierstrass substitution]]

* [[Euler substitution]]

==References==

{{Reflist}}

{{Integrals}}

[[Category:Integral calculus]]

[[Category:Theorems in mathematical analysis]]

[[Category:Theorems in calculus]]