Q17P - Mathematical Methods in Physical Sciences

Question

Evaluate $\mathbf{\int }{\mathbf{e}}^{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{ib}\mathbf{)}\mathbf{x}}\mathbf{dx}$and take real and imaginary parts to show that:

$\mathbf{\int }{\mathbf{e}}^{\mathbf{ax}}\mathbf{cos}\mathbf{ }\mathbf{b}\mathbf{ }\mathbf{xdx}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{ax}}\mathbf{(}\mathbf{a}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{b}\mathbf{ }\mathbf{x}\mathbf{+}\mathbf{b}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{b}\mathbf{ }\mathbf{x}\mathbf{)}}{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

Step-by-Step Solution

Answer

The result of the evaluation of $\int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{b} \sin \mathrm{b} \mathrm{x})+{\mathrm{ie}}^{\mathrm{ax}}\left(\mathrm{a} \sin \mathrm{b} \mathrm{x}-\mathrm{b} \cos \mathrm{bx} )\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}$and the function $\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b} \mathrm{xdx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{b} \sin \mathrm{b} \mathrm{x})}{{\mathrm{a}}^2+{\mathrm{b}}^2}$ has been showed.

Step 1: Given Information.

The given expression is $\int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}$ .

Step 2: Meaning of rectangular form.

Representing the complex number in rectangular form means writing the given complex number in the form of $\mathbf{x}\mathbf{+}\mathbf{iy}$, in which is the real part and y is the imaginary part.

Step 3: Evaluate.

The given question is $\int e^{(a+ib)x}dx$.

$$\begin{gathered} \int e^{(a+ib)x}dx=\frac{e^{{}^{(a+ib)x}}}{a+ib} \\ \int e^{(a+ib)x}dx=\frac{e^{ax}\cdot e^{ibx}}{a+ib} \\ \int e^{(a+ib)x}dx=\frac{e^{ax}[\cos b x+i\sin b x]}{a+ib} \end{gathered}$$

Multiply the numerator and the denominator with the complex conjugate.

$$\begin{gathered} \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}\left[\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{i} \sin \mathrm{b} \mathrm{x}\right]}{{\mathrm{a}}^2+{\mathrm{b}}^2}\times \frac{\mathrm{a}-\mathrm{ib}}{\mathrm{a}-\mathrm{ib}} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}\left[\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{ai} \sin \mathrm{b} \mathrm{x}-\mathrm{ib} \cos \mathrm{bx}+\mathrm{b} \sin \mathrm{b} \mathrm{x}\right]}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}\left(\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{i} \sin \mathrm{b} \mathrm{x}\right)+{\mathrm{ie}}^{\mathrm{ax}}\left(\mathrm{a} \sin \mathrm{b} \mathrm{x}-\mathrm{b} \cos \mathrm{bx} \right)}{{\mathrm{a}}^2+{\mathrm{b}}^2} \end{gathered}$$........(1)

Step 4: Simplify.

$$\begin{gathered} \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{ax}}\cdot {\mathrm{e}}^{\mathrm{ibx}}\mathrm{dx} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{ax}}(\cos \mathrm{b} \mathrm{x}+\mathrm{i} \sin \mathrm{b} \mathrm{x})\mathrm{dx} \\ \int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b} \mathrm{x} \mathrm{dx}+\mathrm{i}\int {\mathrm{e}}^{\mathrm{ax}}\sin \mathrm{b} \mathrm{x} \mathrm{dx} .......\left(2\right) \end{gathered}$$

Using equation (1) and (2).

$\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b} \mathrm{x} \mathrm{dx}+\mathrm{i}\int {\mathrm{e}}^{\mathrm{ax}}\sin \mathrm{b} \mathrm{x} \mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{b} \sin \mathrm{b} \mathrm{x})+{\mathrm{ie}}^{\mathrm{ax}}(\mathrm{a} \sin \mathrm{b} \mathrm{x}-\mathrm{b} \cos \mathrm{b} \mathrm{x})}{{\mathrm{a}}^2+{\mathrm{b}}^2}$

 

Equate the real part on both sides.

$\int e^{ax}cos b x dx=\frac{e^{ax}(a \cos b x+b \sin b x)}{a^2+b^2}$

 

Therefore, the evaluation of $\int {\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{b} \sin \mathrm{b} \mathrm{x})+{\mathrm{ie}}^{\mathrm{ax}}\left(\mathrm{a} \sin \mathrm{b} \mathrm{x}-\mathrm{b} \cos \mathrm{bx} )\right)}{{\mathrm{a}}^2+{\mathrm{b}}^2}$and the function $\int {\mathrm{e}}^{\mathrm{ax}}\cos \mathrm{b} \mathrm{xdx}=\frac{{\mathrm{e}}^{\mathrm{ax}}(\mathrm{a} \cos \mathrm{b} \mathrm{x}+\mathrm{b} \sin \mathrm{b} \mathrm{x})}{{\mathrm{a}}^2+{\mathrm{b}}^2}$ has been showed.