Calculate the improper integral: integral from 0 to ()

Calculate the improper integral: $$ \int_0^\infty e^{-ax}\sin(bx) \, dx $$

This is the companion integral to the one we just solved. It combines exponential decay with sine oscillation.

Step 1: Express as a limit

$$\int_0^\infty e^{-ax}\sin(bx) \, dx = \lim_{R \to \infty} \int_0^R e^{-ax}\sin(bx) \, dx$$

Step 2: Find the antiderivative using integration by parts

Let's denote:

$$I = \int e^{-ax}\sin(bx) \, dx$$

First integration by parts:

Let $u = \sin(bx)$ and $dv = e^{-ax} \, dx$

Then: $du = b\cos(bx) \, dx$ and $v = -\frac{1}{a}e^{-ax}$

$$I = -\frac{1}{a}e^{-ax}\sin(bx) - \int -\frac{1}{a}e^{-ax} \cdot b\cos(bx) \, dx$$

$$= -\frac{1}{a}e^{-ax}\sin(bx) + \frac{b}{a}\int e^{-ax}\cos(bx) \, dx$$

Second integration by parts:

For $\int e^{-ax}\cos(bx) \, dx$, let $u = \cos(bx)$ and $dv = e^{-ax} \, dx$

Then: $du = -b\sin(bx) \, dx$ and $v = -\frac{1}{a}e^{-ax}$

$$\int e^{-ax}\cos(bx) \, dx = -\frac{1}{a}e^{-ax}\cos(bx) - \int -\frac{1}{a}e^{-ax} \cdot (-b\sin(bx)) \, dx$$

$$= -\frac{1}{a}e^{-ax}\cos(bx) - \frac{b}{a}\int e^{-ax}\sin(bx) \, dx$$

$$= -\frac{1}{a}e^{-ax}\cos(bx) - \frac{b}{a}I$$

Substitute back:

$$I = -\frac{1}{a}e^{-ax}\sin(bx) + \frac{b}{a}\left[-\frac{1}{a}e^{-ax}\cos(bx) - \frac{b}{a}I\right]$$

$$= -\frac{1}{a}e^{-ax}\sin(bx) - \frac{b}{a^2}e^{-ax}\cos(bx) - \frac{b^2}{a^2}I$$

Solve for $I$:

$$I + \frac{b^2}{a^2}I = -\frac{1}{a}e^{-ax}\sin(bx) - \frac{b}{a^2}e^{-ax}\cos(bx)$$

$$I\left(1 + \frac{b^2}{a^2}\right) = e^{-ax}\left(-\frac{1}{a}\sin(bx) - \frac{b}{a^2}\cos(bx)\right)$$

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

$$I = \frac{e^{-ax}}{a^2 + b^2}(-a\sin(bx) - b\cos(bx))$$

Step 3: Evaluate the definite integral

$$\int_0^R e^{-ax}\sin(bx) \, dx = \left[\frac{e^{-ax}(-a\sin(bx) - b\cos(bx))}{a^2 + b^2}\right]_0^R$$

At $x = R$:

$$\frac{e^{-aR}(-a\sin(bR) - b\cos(bR))}{a^2 + b^2}$$

At $x = 0$:

$$\frac{e^0(-a\sin 0 - b\cos 0)}{a^2 + b^2} = \frac{1 \cdot (0 - b \cdot 1)}{a^2 + b^2} = \frac{-b}{a^2 + b^2}$$

Therefore:

$$\int_0^R e^{-ax}\sin(bx) \, dx = \frac{e^{-aR}(-a\sin(bR) - b\cos(bR))}{a^2 + b^2} - \frac{-b}{a^2 + b^2}$$

$$= \frac{e^{-aR}(-a\sin(bR) - b\cos(bR)) + b}{a^2 + b^2}$$

Step 4: Take the limit

$$\int_0^\infty e^{-ax}\sin(bx) \, dx = \lim_{R \to \infty} \frac{e^{-aR}(-a\sin(bR) - b\cos(bR)) + b}{a^2 + b^2}$$

Case 1: $a > 0$

As $R \to \infty$: $e^{-aR} \to 0$

Since $|\sin(bR)| \leq 1$ and $|\cos(bR)| \leq 1$:

$$|e^{-aR}(-a\sin(bR) - b\cos(bR))| \leq e^{-aR}(a + b) \to 0$$

Therefore:

$$\lim_{R \to \infty} e^{-aR}(-a\sin(bR) - b\cos(bR)) = 0$$

Thus:

$$\int_0^\infty e^{-ax}\sin(bx) \, dx = \frac{0 + b}{a^2 + b^2} = \frac{b}{a^2 + b^2}$$

Case 2: $a \leq 0$

If $a = 0$: the integral becomes $\int_0^\infty \sin(bx) \, dx$, which diverges (by oscillation).

If $a < 0$: $e^{-aR} = e^{|a|R} \to +\infty$, so the limit doesn't exist and the integral diverges.

Final Answer

$$\boxed{\int_0^\infty e^{-ax}\sin(bx) \, dx = \begin{cases} \dfrac{b}{a^2 + b^2} & \text{if } a > 0 \quad \text{(converges)}\\[10pt] \text{divergent} & \text{if } a \leq 0 \end{cases}}$$

Special cases

Case 1: $a = 1, b = 0$

$$\int_0^\infty e^{-x}\sin(0) \, dx = \int_0^\infty 0 \, dx = 0 = \frac{0}{1 + 0}$$

Case 2: $a = 1, b = 1$

$$\int_0^\infty e^{-x}\sin(x) \, dx = \frac{1}{1 + 1} = \frac{1}{2}$$

Case 3: $a = 2, b = 3$

$$\int_0^\infty e^{-2x}\sin(3x) \, dx = \frac{3}{4 + 9} = \frac{3}{13}$$

Case 4: $a = 1, b = 2$

$$\int_0^\infty e^{-x}\sin(2x) \, dx = \frac{2}{1 + 4} = \frac{2}{5}$$

Relationship with the cosine integral

We now have the pair:

$$\int_0^\infty e^{-ax}\cos(bx) \, dx = \frac{a}{a^2 + b^2} \quad \text{(for } a > 0\text{)}$$

$$\int_0^\infty e^{-ax}\sin(bx) \, dx = \frac{b}{a^2 + b^2} \quad \text{(for } a > 0\text{)}$$

Beautiful property:

$$\left(\frac{a}{a^2 + b^2}\right)^2 + \left(\frac{b}{a^2 + b^2}\right)^2 = \frac{a^2 + b^2}{(a^2 + b^2)^2} = \frac{1}{a^2 + b^2}$$

This Pythagorean relationship reflects the orthogonality of sine and cosine!

Complex exponential approach (alternative method)

Using Euler's formula: $\sin(bx) = \text{Im}(e^{ibx})$

$$\int_0^\infty e^{-ax}\sin(bx) \, dx = \text{Im}\left(\int_0^\infty e^{-ax}e^{ibx} \, dx\right)$$

$$= \text{Im}\left(\int_0^\infty e^{-(a-ib)x} \, dx\right)$$

$$= \text{Im}\left(\left[\frac{e^{-(a-ib)x}}{-(a-ib)}\right]_0^\infty\right)$$

For $a > 0$:

$$= \text{Im}\left(\frac{0 - 1}{-(a-ib)}\right) = \text{Im}\left(\frac{1}{a-ib}\right)$$

Multiply by conjugate:

$$= \text{Im}\left(\frac{1}{a-ib} \cdot \frac{a+ib}{a+ib}\right) = \text{Im}\left(\frac{a+ib}{a^2+b^2}\right) = \frac{b}{a^2+b^2}$$

This confirms our answer! ✓

Verification using derivative

Let's verify the antiderivative. If:

$$F(x) = \frac{e^{-ax}(-a\sin(bx) - b\cos(bx))}{a^2 + b^2}$$

Then using the product rule and chain rule:

$$F'(x) = \frac{1}{a^2+b^2}\left[(-a)e^{-ax}(-a\sin(bx) - b\cos(bx)) + e^{-ax}(-ab\cos(bx) + b^2\sin(bx))\right]$$

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

$$= \frac{e^{-ax}}{a^2+b^2}(a^2 + b^2)\sin(bx) = e^{-ax}\sin(bx)$$

Correct! ✓

Laplace transform interpretation

The Laplace transform of $\sin(bt)$ is:

$$\mathcal{L}\{\sin(bt)\} = \int_0^\infty e^{-st}\sin(bt) \, dt = \frac{b}{s^2 + b^2}$$

This is exactly our result with $a = s$. This is a fundamental Laplace transform!

Physical interpretation

The function $e^{-at}\sin(bt)$ represents:

• Damped harmonic oscillation
• Initial amplitude starts from 0 (since $\sin(0) = 0$)
• Grows initially, then decays exponentially
• Used in circuit theory, mechanical vibrations, etc.

The integral $\frac{b}{a^2+b^2}$ represents the total "energy" or "impulse" over all time.

Comparison: cosine vs. sine

$$\begin{array}{c|c|c} \text{Function} & \text{Initial value} & \text{Integral} \\ \hline e^{-ax}\cos(bx) & 1 \text{ (at } x=0\text{)} & \dfrac{a}{a^2+b^2} \\[10pt] e^{-ax}\sin(bx) & 0 \text{ (at } x=0\text{)} & \dfrac{b}{a^2+b^2} \end{array}$$

The cosine starts at maximum, the sine starts at zero - this is reflected in the numerators!

Linear combination

For any constants $\alpha$ and $\beta$:

$$\int_0^\infty e^{-ax}[\alpha\cos(bx) + \beta\sin(bx)] \, dx = \frac{\alpha a + \beta b}{a^2 + b^2}$$

This can be written as:

$$= \frac{1}{a^2+b^2}\sqrt{(\alpha a)^2 + (\beta b)^2} \cdot \cos(\phi)$$

where $\phi$ is an appropriate phase angle.

Summary

1. The integral combines exponential decay with sine oscillation
2. Integration by parts (twice) gives the antiderivative
3. Convergence requires $a > 0$ (exponential decay)
4. When convergent: $\int_0^\infty e^{-ax}\sin(bx) \, dx = \frac{b}{a^2 + b^2}$
5. Companion result: $\int_0^\infty e^{-ax}\cos(bx) \, dx = \frac{a}{a^2 + b^2}$
6. These are fundamental Laplace transforms

Key lessons:

• Sine and cosine integrals with exponential decay form a complementary pair
• The numerators ($a$ vs $b$) reflect initial conditions at $x = 0$
• Both require $a > 0$ for convergence
• The complex exponential method elegantly unifies both results
• These formulas are central to Laplace transform theory
• The results satisfy a Pythagorean relationship
• Integration by parts is the key technique for products of exponentials and trig functions