Q18P - Mathematical Methods in Physical Sciences

Question

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

$\mathbf{\int }{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{b}\mathit{x}\mathit{d}\mathit{x}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{ax}}\mathbf{ }\left(a \sin \mathrm{bx}-b \mathrm{cod} \mathrm{bx}\right)}{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

Step-by-Step Solution

Answer

The result of the evaluation of 

$\int e^{\left(a+ib\right)x} dx=\frac{e^{ax}\left(a \cos bx+b \sin bx\right)+i{e}^{ax}\left(a \sin bx+b cos bx\right)}{a^2+b^2}$and the function $\int e^{ax} \sin bxdx=\frac{e^{ax}\left(a \sin bx-b \cos bx\right)}{a^2+b^2}$has been showed.

Step 1: Given Information.

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

Step 2: Meaning of rectangular form.

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

Step 3: Evaluate.

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

$$\begin{gathered} \int {\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} dx=\frac{{\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}}}{a+ib} \\ \int {\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} dx=\frac{{\mathrm{e}}^{\mathrm{ax}} .{\mathrm{e}}^{\mathrm{ibx}}}{a+ib} \\ \int {\mathrm{e}}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} dx=\frac{{\mathrm{e}}^{ax}\left[\cos \mathrm{bx}+\mathrm{i} \sin \mathrm{bx}\right]}{a+ib} \end{gathered}$$

 

Multiply the numerator and the denominator with the complex conjugate of 

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

Step 4: Simplify.

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

 

Using equation (1) and (2).

$\int e^{ax} \cos bxdx+i\int e^{ax} \sin bxdx =\frac{e^{ax}\left(a \cos bx+b \sin bx\right)+i{e}^{ax}\left(a \sin bx-b \cos bx\right)}{a^2+b^2}$

 

Equate the imaginary part on both sides.

$\int e^{ax} \sin bxdx=\frac{e^{ax}\left(a \sin bx-b cos bx\right)}{a^2+b^2}$

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