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}} and e − i x {\displaystyle e^{-ix}} 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 expression involving trigonometric functions.[1]
Euler's formula [ edit ] Euler's formula states that [2]
e i x = cos x + i sin x . {\displaystyle e^{ix}=\cos x+i\,\sin x.} Substituting − x {\displaystyle -x} for x {\displaystyle x} gives the equation
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 [ edit ] First example [ edit ] 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 . Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:
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 [ edit ] 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 [ edit ] 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 , we know that
∫ 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 [ edit ] 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}} , the result is the integral of a 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 [ edit ] References [ edit ]