The average value of a function f(x) on the interval [0

The average value of a function $f(x)$ on the interval $[0, \infty)$ is defined as $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(x) \, dx $$ Find the average value of the function $f(x) = \arctan x$ on the interval $[0, \infty)$.

Step 1: Set up the formula

We need to calculate:

$$\bar{f} = \lim_{t \to \infty} \frac{1}{t} \int_0^t \arctan x \, dx$$

Step 2: Find $\int \arctan x \, dx$

We use integration by parts.

Let $u = \arctan x$ and $dv = dx$.

Then: $du = \frac{1}{1+x^2} \, dx$ and $v = x$

By integration by parts:

$$\int \arctan x \, dx = x \arctan x - \int x \cdot \frac{1}{1+x^2} \, dx$$

For the remaining integral, let $w = 1 + x^2$, so $dw = 2x \, dx$:

$$\int \frac{x}{1+x^2} \, dx = \frac{1}{2}\int \frac{dw}{w} = \frac{1}{2}\ln|w| + C = \frac{1}{2}\ln(1+x^2) + C$$

Therefore:

$$\int \arctan x \, dx = x \arctan x - \frac{1}{2}\ln(1+x^2) + C$$

Step 3: Evaluate the definite integral

$$\int_0^t \arctan x \, dx = \left[x \arctan x - \frac{1}{2}\ln(1+x^2)\right]_0^t$$

At $x = t$:

$$t \arctan t - \frac{1}{2}\ln(1+t^2)$$

At $x = 0$:

$$0 \cdot \arctan 0 - \frac{1}{2}\ln(1) = 0 - 0 = 0$$

Therefore:

$$\int_0^t \arctan x \, dx = t \arctan t - \frac{1}{2}\ln(1+t^2)$$

Step 4: Calculate the limit

$$\bar{f} = \lim_{t \to \infty} \frac{1}{t}\left[t \arctan t - \frac{1}{2}\ln(1+t^2)\right]$$

$$= \lim_{t \to \infty} \left[\arctan t - \frac{\ln(1+t^2)}{2t}\right]$$

First term:

$$\lim_{t \to \infty} \arctan t = \frac{\pi}{2}$$

Second term:

$$\lim_{t \to \infty} \frac{\ln(1+t^2)}{2t}$$

For large $t$: $\ln(1+t^2) \approx \ln(t^2) = 2\ln t$

So:

$$\lim_{t \to \infty} \frac{\ln(1+t^2)}{2t} = \lim_{t \to \infty} \frac{2\ln t}{2t} = \lim_{t \to \infty} \frac{\ln t}{t}$$

By L'Hôpital's rule:

$$\lim_{t \to \infty} \frac{\ln t}{t} = \lim_{t \to \infty} \frac{1/t}{1} = \lim_{t \to \infty} \frac{1}{t} = 0$$

Combining:

$$\bar{f} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

$$\boxed{\bar{f} = \frac{\pi}{2}}$$

Step 5: Numerical value

$$\bar{f} = \frac{\pi}{2} \approx 1.5708$$

$$\boxed{\bar{f} \approx 1.571}$$

Interpretation

The average value $\frac{\pi}{2}$ equals the limiting value of $\arctan x$ as $x \to \infty$.

This makes intuitive sense because:

• $\arctan x$ increases from 0 to $\frac{\pi}{2}$ as $x$ goes from 0 to $\infty$
• $\arctan x$ approaches $\frac{\pi}{2}$ asymptotically
• For large intervals $[0, t]$, the function spends most of its "time" near the value $\frac{\pi}{2}$

Verification of the calculation

Let's verify the limit more carefully:

$$\lim_{t \to \infty} \frac{\ln(1+t^2)}{2t}$$

Write $\ln(1+t^2) = \ln(t^2(1/t^2 + 1)) = \ln t^2 + \ln(1 + 1/t^2) = 2\ln t + \ln(1 + 1/t^2)$

Therefore:

$$\frac{\ln(1+t^2)}{2t} = \frac{2\ln t + \ln(1+1/t^2)}{2t} = \frac{\ln t}{t} + \frac{\ln(1+1/t^2)}{2t}$$

As $t \to \infty$:

• $\frac{\ln t}{t} \to 0$ (by L'Hôpital)
• $\ln(1+1/t^2) \to \ln 1 = 0$, so $\frac{\ln(1+1/t^2)}{2t} \to 0$

Therefore:

$$\lim_{t \to \infty} \frac{\ln(1+t^2)}{2t} = 0 \quad \checkmark$$

Summary

Average value of $f(x) = \arctan x$ on $[0, \infty)$:

$$\boxed{\bar{f} = \frac{\pi}{2} \approx 1.571}$$

Key steps:

1. Use definition: $\bar{f} = \lim_{t \to \infty} \frac{1}{t} \int_0^t f(x) \, dx$
2. Integration by parts: $\int \arctan x \, dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C$
3. Evaluate: $\int_0^t \arctan x \, dx = t\arctan t - \frac{1}{2}\ln(1+t^2)$
4. Take limit: $\lim_{t \to \infty} \left[\arctan t - \frac{\ln(1+t^2)}{2t}\right] = \frac{\pi}{2} - 0$

Key insights:

• The average value equals the horizontal asymptote $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$
• The logarithmic term vanishes: $\frac{\ln(1+t^2)}{2t} \to 0$
• L'Hôpital's rule shows $\frac{\ln t}{t} \to 0$
• Beautiful result: the average is exactly $\frac{\pi}{2}$
• This generalizes the concept of average value to infinite intervals