Determine the sign of the integral: I = integral from 1

Determine the sign of the integral: $$ I = \int_1^e \frac{\arctan(\ln x)}{x} \, dx $$

Method 1: Sign analysis of the integrand

Step 1: Analyze $\ln x$ on $[1, e]$

For $x \in [1, e]$:

• At $x = 1$: $\ln 1 = 0$
• At $x = e$: $\ln e = 1$
• For $x \in (1, e)$: $0 < \ln x < 1$

Therefore: $\ln x \in [0, 1]$ for $x \in [1, e]$

Step 2: Analyze $\arctan(\ln x)$

Since $\ln x \geq 0$ on $[1, e]$ and $\arctan$ is an odd function with $\arctan t > 0$ for $t > 0$:

For $x \in [1, e]$:

$$\arctan(\ln x) \geq 0$$

Moreover, $\arctan(\ln x) > 0$ for $x \in (1, e]$ since $\ln x > 0$ there.

Specifically:

• At $x = 1$: $\arctan(\ln 1) = \arctan(0) = 0$
• At $x = e$: $\arctan(\ln e) = \arctan(1) = \frac{\pi}{4} > 0$
• For $x \in (1, e)$: $\arctan(\ln x) > 0$

Step 3: Analyze $\frac{1}{x}$

For $x \in [1, e]$:

$$\frac{1}{x} > 0$$

Step 4: Analyze the integrand $f(x) = \frac{\arctan(\ln x)}{x}$

For $x \in [1, e]$:

$$f(x) = \frac{\arctan(\ln x)}{x} \geq 0$$

Moreover:

• At $x = 1$: $f(1) = \frac{\arctan(0)}{1} = 0$
• For $x \in (1, e]$: $f(x) > 0$

Step 5: Determine the sign of the integral

Since $f(x) \geq 0$ on $[1, e]$ and $f(x) > 0$ on $(1, e]$:

$$I = \int_1^e f(x) \, dx > 0$$

$$\boxed{I > 0 \quad \text{(The integral is positive)}}$$

Method 2: Direct calculation

We can also compute the integral explicitly using substitution.

Step 1: Substitution

Let $u = \ln x$, so $du = \frac{1}{x} \, dx$.

When $x = 1$: $u = \ln 1 = 0$

When $x = e$: $u = \ln e = 1$

Therefore:

$$I = \int_1^e \frac{\arctan(\ln x)}{x} \, dx = \int_0^1 \arctan u \, du$$

Step 2: Integrate $\arctan u$

Using integration by parts with $v = \arctan u$ and $dw = du$:

$dv = \frac{1}{1+u^2} \, du$ and $w = u$

$$\int \arctan u \, du = u \arctan u - \int \frac{u}{1+u^2} \, du$$

For the remaining integral, let $s = 1 + u^2$, so $ds = 2u \, du$:

$$\int \frac{u}{1+u^2} \, du = \frac{1}{2}\ln(1+u^2) + C$$

Therefore:

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

Step 3: Evaluate the definite integral

$$I = \left[u \arctan u - \frac{1}{2}\ln(1+u^2)\right]_0^1$$

At $u = 1$:

$$1 \cdot \arctan 1 - \frac{1}{2}\ln(1+1) = \frac{\pi}{4} - \frac{1}{2}\ln 2$$

At $u = 0$:

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

Therefore:

$$I = \frac{\pi}{4} - \frac{\ln 2}{2}$$

$$\boxed{I = \frac{\pi}{4} - \frac{\ln 2}{2}}$$

Step 4: Determine the sign

$$I = \frac{\pi}{4} - \frac{\ln 2}{2} \approx 0.7854 - 0.3466 \approx 0.439$$

Since $\frac{\pi}{4} \approx 0.785$ and $\frac{\ln 2}{2} \approx 0.347$:

$$I \approx 0.439 > 0$$

This confirms our result: $I > 0$ ✓

Verification that $\frac{\pi}{4} > \frac{\ln 2}{2}$

We need to verify: $\frac{\pi}{4} > \frac{\ln 2}{2}$

Equivalently: $\frac{\pi}{2} > \ln 2$

Or: $e^{\pi/2} > 2$

Since $e^{\pi/2} \approx e^{1.571} \approx 4.81 > 2$, this is true. ✓

Alternatively:

$$\frac{\pi}{4} \approx 0.7854 \quad \text{and} \quad \frac{\ln 2}{2} = \frac{0.6931}{2} \approx 0.3466$$

Clearly $0.7854 > 0.3466$ ✓

Summary

Sign of the integral:

$$\boxed{I = \int_1^e \frac{\arctan(\ln x)}{x} \, dx > 0 \quad \text{(POSITIVE)}}$$

Exact value:

$$\boxed{I = \frac{\pi}{4} - \frac{\ln 2}{2} \approx 0.439}$$

Key reasons why $I > 0$:

1. On $[1, e]$: $\ln x \geq 0$, so $\arctan(\ln x) \geq 0$
2. On $[1, e]$: $\frac{1}{x} > 0$
3. Therefore: integrand $\frac{\arctan(\ln x)}{x} \geq 0$ on $[1, e]$
4. The integrand is strictly positive on $(1, e]$
5. Thus the integral is positive

Key insights:

• Sign analysis: integrand is non-negative, and positive on most of the interval
• Substitution $u = \ln x$ simplifies to $\int_0^1 \arctan u \, du$
• Exact value: $\frac{\pi}{4} - \frac{\ln 2}{2} \approx 0.439$
• The integral is definitely positive
• Integration by parts gives exact closed form