Use of complex numbers to evaluate integrals
In integral calculus , Euler's formula for complex numbers may be used to evaluate integrals involving trigonometric functions . Using Euler's formula, any trigonometric function may be written in terms of complex exponential functions, namely
e
i
x
{\displaystyle e^{ix}}
e
−
i
x
{\displaystyle e^{-ix}}
trigonometric identities or integration by parts , and is sufficiently powerful to integrate any rational expression involving trigonometric functions.
Euler's formula states that [ 1]
e
i
x
=
cos
x
+
i
sin
x
.
{\displaystyle e^{ix}=\cos x+i\,\sin x.}
Substituting
−
x
{\displaystyle -x}
x
{\displaystyle x}
e
−
i
x
=
cos
x
−
i
sin
x
{\displaystyle e^{-ix}=\cos x-i\,\sin x}
because cosine is an even function and sine is odd. These two equations can be solved for the sine and cosine to give
cos
x
=
e
i
x
+
e
−
i
x
2
and
sin
x
=
e
i
x
−
e
−
i
x
2
i
.
{\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}\quad {\text{and}}\quad \sin x={\frac {e^{ix}-e^{-ix}}{2i}}.}
Examples
First example
Consider the integral
∫
cos
2
x
d
x
.
{\displaystyle \int \cos ^{2}x\,dx.}
The standard approach to this integral is to use a half-angle formula to simplify the integrand. We can use Euler's identity instead:
∫
cos
2
x
d
x
=
∫
(
e
i
x
+
e
−
i
x
2
)
2
d
x
=
1
4
∫
(
e
2
i
x
+
2
+
e
−
2
i
x
)
d
x
{\displaystyle {\begin{aligned}\int \cos ^{2}x\,dx\,&=\,\int \left({\frac {e^{ix}+e^{-ix}}{2}}\right)^{2}dx\\[6pt]&=\,{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx\end{aligned}}}
At this point, it would be possible to change back to real numbers using the formula e 2ix + e −2ix = 2 cos 2x
1
4
∫
(
e
2
i
x
+
2
+
e
−
2
i
x
)
d
x
=
1
4
(
e
2
i
x
2
i
+
2
x
−
e
−
2
i
x
2
i
)
+
C
=
1
4
(
2
x
+
sin
2
x
)
+
C
.
{\displaystyle {\begin{aligned}{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx&={\frac {1}{4}}\left({\frac {e^{2ix}}{2i}}+2x-{\frac {e^{-2ix}}{2i}}\right)+C\\[6pt]&={\frac {1}{4}}\left(2x+\sin 2x\right)+C.\end{aligned}}}
Second example
Consider the integral
∫
sin
2
x
cos
4
x
d
x
.
{\displaystyle \int \sin ^{2}x\cos 4x\,dx.}
This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless:
∫
sin
2
x
cos
4
x
d
x
=
∫
(
e
i
x
−
e
−
i
x
2
i
)
2
(
e
4
i
x
+
e
−
4
i
x
2
)
d
x
=
−
1
8
∫
(
e
2
i
x
−
2
+
e
−
2
i
x
)
(
e
4
i
x
+
e
−
4
i
x
)
d
x
=
−
1
8
∫
(
e
6
i
x
−
2
e
4
i
x
+
e
2
i
x
+
e
−
2
i
x
−
2
e
−
4
i
x
+
e
−
6
i
x
)
d
x
.
{\displaystyle {\begin{aligned}\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]&=-{\frac {1}{8}}\int \left(e^{2ix}-2+e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right)dx\\[6pt]&=-{\frac {1}{8}}\int \left(e^{6ix}-2e^{4ix}+e^{2ix}+e^{-2ix}-2e^{-4ix}+e^{-6ix}\right)dx.\end{aligned}}}
At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there.
Either method gives
∫
sin
2
x
cos
4
x
d
x
=
−
1
24
sin
6
x
+
1
8
sin
4
x
−
1
8
sin
2
x
+
C
.
{\displaystyle \int \sin ^{2}x\cos 4x\,dx=-{\frac {1}{24}}\sin 6x+{\frac {1}{8}}\sin 4x-{\frac {1}{8}}\sin 2x+C.}
Using real parts
In addition to Euler's identity, it can be helpful to make judicious use of the real parts of complex expressions. For example, consider the integral
∫
e
x
cos
x
d
x
.
{\displaystyle \int e^{x}\cos x\,dx.}
Since cos x is the real part of e ix
∫
e
x
cos
x
d
x
=
Re
∫
e
x
e
i
x
d
x
.
{\displaystyle \int e^{x}\cos x\,dx=\operatorname {Re} \int e^{x}e^{ix}\,dx.}
The integral on the right is easy to evaluate:
∫
e
x
e
i
x
d
x
=
∫
e
(
1
+
i
)
x
d
x
=
e
(
1
+
i
)
x
1
+
i
+
C
.
{\displaystyle \int e^{x}e^{ix}\,dx=\int e^{(1+i)x}\,dx={\frac {e^{(1+i)x}}{1+i}}+C.}
Thus:
∫
e
x
cos
x
d
x
=
Re
(
e
(
1
+
i
)
x
1
+
i
)
+
C
=
e
x
Re
(
e
i
x
1
+
i
)
+
C
=
e
x
Re
(
e
i
x
(
1
−
i
)
2
)
+
C
=
e
x
cos
x
+
sin
x
2
+
C
.
{\displaystyle {\begin{aligned}\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{aligned}}}
Fractions
In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral
∫
1
+
cos
2
x
cos
x
+
cos
3
x
d
x
.
{\displaystyle \int {\frac {1+\cos ^{2}x}{\cos x+\cos 3x}}\,dx.}
Using Euler's identity, this integral becomes
1
2
∫
6
+
e
2
i
x
+
e
−
2
i
x
e
i
x
+
e
−
i
x
+
e
3
i
x
+
e
−
3
i
x
d
x
.
{\displaystyle {\frac {1}{2}}\int {\frac {6+e^{2ix}+e^{-2ix}}{e^{ix}+e^{-ix}+e^{3ix}+e^{-3ix}}}\,dx.}
If we now make the substitution
u
=
e
i
x
{\displaystyle u=e^{ix}}
rational function :
−
i
2
∫
1
+
6
u
2
+
u
4
1
+
u
2
+
u
4
+
u
6
d
u
.
{\displaystyle -{\frac {i}{2}}\int {\frac {1+6u^{2}+u^{4}}{1+u^{2}+u^{4}+u^{6}}}\,du.}
One may proceed using partial fraction decomposition .
See also
References
![{\displaystyle {\begin{aligned}\int \cos ^{2}x\,dx\,&=\,\int \left({\frac {e^{ix}+e^{-ix}}{2}}\right)^{2}dx\\[6pt]&=\,{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/477bcd8b7e0424bbab7594999c9a5157a119d1f4)
![{\displaystyle {\begin{aligned}{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx&={\frac {1}{4}}\left({\frac {e^{2ix}}{2i}}+2x-{\frac {e^{-2ix}}{2i}}\right)+C\\[6pt]&={\frac {1}{4}}\left(2x+\sin 2x\right)+C.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/094b4102f3fcb40250e8417ebff895de2aed2b46)
![{\displaystyle {\begin{aligned}\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]&=-{\frac {1}{8}}\int \left(e^{2ix}-2+e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right)dx\\[6pt]&=-{\frac {1}{8}}\int \left(e^{6ix}-2e^{4ix}+e^{2ix}+e^{-2ix}-2e^{-4ix}+e^{-6ix}\right)dx.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/803fea23145359a202cb5de0ea639ddbfbc77b0f)
![{\displaystyle {\begin{aligned}\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65e9c7fbcba2dfcd2ca6dab6894ec6bb9e21f128)
