Integration using Euler's formula: Difference between revisions

Revision as of 11:51, 6 November 2008

Latest revision as of 04:50, 12 July 2025

(94 intermediate revisions by 60 users not shown)

Line 1:

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

{{Cleanup|date=August 2008}}

{{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==

Functions containing sine or cosine can be expressed as complex exponentials using

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>

[[Euler's formula]].

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

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

Example: suppose we wanted to integrate:

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

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

==Examples==

Then the cosine function can be expressed in its Euler form: <math>\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}</math>

=== First example ===

: <math>\int e^x \cdot \frac{e^{ix} + e^{-ix}}{2} \, dx</math>

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===

: <math>{1\over 2} \int e^{x(1+i)} + e^{x(1-i)} \, dx</math>

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==

This is far easier to integrate.

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==

Alternatively, we may also take note of real and imaginary portions of complex numbers

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==

Cosine is the real portion of a complex number written in cos x + i sin x form

{{Portal|Mathematics}}

* [[Trigonometric substitution]]

* [[Weierstrass substitution]]

* [[Euler substitution]]

==References==

<math>\int e^x \cos x dx =</math>

{{Reflist}}

{{Integrals}}

<math>\int e^x \mathrm{Re}\{ \cos x + i\cdot \sin x \} dx</math>

<math>\int e^x \mathrm{Re}\{ e^{ix} \} dx</math>

<math>\mathrm{Re}\{ \int e^x e^{ix} dx \}</math>

<math>\mathrm{Re}\{ \int e^{(i+1)x} dx \}</math>

This calculation continues as:

=Re (1/(1+i)) * exp((1+i)*x)

=Re ( 1/2 + i*1/2 ) * exp(x) * (cos (x) +i*sin(x))

=Re 1/2*exp(x)*cos(x)+1/2*i*exp(x)*sin(x)-1/2*i*exp(x)*cos(x)+1/2*exp(x)*sin(x)

=1/2 exp(x)*cos(x) + 1/2 exp(x)*sin(x)

[[Category:Integral calculus]]

[[Category:Theorems in mathematical analysis]]

[[Category:Theorems in calculus]]

{{math-stub}}