Chapter I, Problem 50E

1. (a) If u is a complex-valued function of a real variable, its indefinite integral ${\displaystyle \int u(x)}dx$ is an antiderivative of u. Evaluate

${\displaystyle \int e^{(1+i)x}dx}$

1. (b) By considering the real and imaginary parts of the integral in part (a), evaluate the real integrals

${\displaystyle \int e^x\cos xdx}\text{and}{\displaystyle \int e^x\sin xdx}$

1. (c) Compare with the method used in Example 7.1.4.

To determine

To evaluate: The integral ${\displaystyle \int e^{\left(1+i\right)x}}dx$.

Answer to Problem 50E

The value of the integral is $\frac{1-i}{2}{e}^{\left(1+i\right)x}+c$.

Explanation of Solution

Calculation:

From the exercise 49, it is known that if $F\left(x\right)=e^{\left(1+i\right)x}$, then $F^{\prime }\left(x\right)=\left(1+i\right)e^{\left(1+i\right)x}$.

Therefore, the given integral becomes,

$\begin{array}{c}{\displaystyle \int e^{\left(1+i\right)x}}dx={\displaystyle \int \frac{F^{\prime }\left(x\right)}{1+i}}dx\\ =\frac{1}{1+i}{\displaystyle \int F^{\prime }\left(x\right)}dx\\ =\frac{1}{1+i}F\left(x\right)+c\\ =\frac{1}{1+i}{e}^{\left(1+i\right)x}+c\end{array}$

Simplify further as,

$\begin{array}{c}{\displaystyle \int e^{\left(1+i\right)x}}dx=\frac{1}{1+i}\left(\frac{1-i}{1-i}\right)e^{\left(1+i\right)x}+c\\ =\frac{1-i}{1^2-i^2}{e}^{\left(1+i\right)x}+c\\ =\frac{1-i}{1+1}{e}^{\left(1+i\right)x}+c\\ =\frac{1-i}{2}{e}^{\left(1+i\right)x}+c\end{array}$

Thus, the value of the integral is $\frac{1-i}{2}{e}^{\left(1+i\right)x}+c$.

To determine

To evaluate: The integral ${\displaystyle \int e^x\cos x}dx$ and ${\displaystyle \int e^x\sin x}dx$.

Answer to Problem 50E

The value of the integral is ${\displaystyle \int e^x\cos x}dx=\frac{e^x}{2}\left(\cos x+\sin x\right)$ and ${\displaystyle \int e^x\sin x}dx=\frac{e^x}{2}\left(\sin x-\cos x\right)$.

Explanation of Solution

Calculation:

Rewrite the given integral as,

$\begin{array}{c}{\displaystyle \int e^{\left(1+i\right)x}}dx={\displaystyle \int e^x\cdot e^{ix}}dx\\ ={\displaystyle \int e^x\cdot \left(\cos x+i\sin x\right)}dx\\ ={\displaystyle \int e^x\cos x}dx+i{\displaystyle \int e^x\sin x}dx\end{array}$

Thus, ${\displaystyle \int e^{\left(1+i\right)x}}dx={\displaystyle \int e^x\cos x}dx+i{\displaystyle \int e^x\sin x}dx$.                                                            (1)

From part (a), it is known that ${\displaystyle \int e^{\left(1+i\right)x}}dx=\frac{1-i}{2}{e}^{\left(1+i\right)x}+c$.

Simplify this as,

$\begin{array}{c}\frac{1-i}{2}{e}^{\left(1+i\right)x}=\frac{e^{\left(1+i\right)x}}{2}-i\frac{e^{\left(1+i\right)x}}{2}\\ =\frac{1}{2}\left[e^x\left(\cos x+i\sin x\right)-i{e}^x\left(\cos x+i\sin x\right)\right]\\ =\frac{e^x}{2}\left[\cos x+i\sin x-i\cos x+\sin x\right]\\ =\frac{e^x}{2}\left[\cos x+\sin x+i\left(\sin x-\cos x\right)\right]\end{array}$

Thus, $\frac{1-i}{2}{e}^{\left(1+i\right)x}=\frac{e^x}{2}\left(\cos x+\sin x\right)+\frac{i{e}^x}{2}\left(\sin x-\cos x\right)$.                                         (2)

Compare the equations (1) and (2), it is observed that ${\displaystyle \int e^x\cos x}dx=\frac{e^x}{2}\left(\cos x+\sin x\right)$ and ${\displaystyle \int e^x\sin x}dx=\frac{e^x}{2}\left(\sin x-\cos x\right)$.