Calculate the improper integral: integral from 0 to ()

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

This integral combines exponential decay with trigonometric oscillation. The convergence depends on the parameter $a$.

Step 1: Express as a limit

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

Step 2: Find the antiderivative using integration by parts

This requires integration by parts twice. Let's denote:

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

First integration by parts:

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

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

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

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

Second integration by parts:

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

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

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

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

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

Substitute back:

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

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

Solve for $I$:

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

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

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

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

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

Step 3: Evaluate the definite integral

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

At $x = R$:

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

At $x = 0$:

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

Therefore:

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

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

Step 4: Take the limit

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

Case 1: $a > 0$

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

Note that $|\cos(bR)| \leq 1$ and $|\sin(bR)| \leq 1$, so:

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

Therefore:

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

Thus:

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

Case 2: $a \leq 0$

If $a = 0$: the integral becomes $\int_0^\infty \cos(bx) \, dx$, which diverges (as we saw earlier).

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

Final Answer

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

Special cases

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

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

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

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

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

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

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

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

Companion integral (sine version)

By a similar method, we can show:

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

Relationship:

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

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

Note that:

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

Complex exponential approach (alternative method)

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

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

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

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

For $a > 0$:

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

Multiply by conjugate:

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

This confirms our answer! ✓

Physical applications

This integral appears in many areas:

1. Laplace Transform:

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

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

This is exactly our result with $a = s$.

2. Damped oscillations:

In physics, $e^{-at}\cos(bt)$ represents a damped harmonic oscillator:

• $e^{-at}$ is the exponential decay (damping)
• $\cos(bt)$ is the oscillation
• The total "energy" (area under curve) is $\frac{a}{a^2 + b^2}$

3. Signal processing:

The integral represents the response of a system to a sinusoidal input with exponential damping.

Why the exponential decay is necessary

Without the factor $e^{-ax}$ (i.e., when $a = 0$), we have:

$$\int_0^\infty \cos(bx) \, dx$$

which diverges by oscillation, as we saw earlier.

The exponential decay $e^{-ax}$ (when $a > 0$):

• "Tames" the oscillation
• Causes the integrand to approach 0 as $x \to \infty$
• Makes the integral converge

The larger $a$ is, the faster the decay and the smaller the integral value.

Behavior as parameters vary

For fixed $b > 0$:

$$\frac{a}{a^2 + b^2} = \frac{1}{a + b^2/a}$$

As $a \to 0^+$: integral $\to 0$ (approaching divergence)

As $a \to \infty$: integral $\to 0$ (very strong damping)

Maximum occurs when $\frac{d}{da}\left[\frac{a}{a^2+b^2}\right] = 0$:

$$\frac{(a^2+b^2) - a(2a)}{(a^2+b^2)^2} = 0 \implies b^2 - a^2 = 0 \implies a = b$$

At $a = b$: $\frac{a}{a^2+a^2} = \frac{1}{2a}$

Summary

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

Key lessons:

• Exponential decay can make oscillating integrals converge
• Integration by parts is essential for products of exponentials and trig functions
• The result has beautiful applications in Laplace transforms and physics
• The complex exponential method provides an elegant alternative derivation
• The parameter $a$ must be positive for convergence
• The formula $\frac{a}{a^2+b^2}$ is fundamental in transform theory