If f(x) = arctan 1/x - 1/(1+x²) and I_n = integral from 0

If $f(x) = \arctan\frac{1}{x} - \frac{1}{1+x^2}$ and $I_n = \int_0^n f(x) \, dx$, $J_n = \int_{-n}^0 f(x) \, dx$, find: $$ \lim_{n \to \infty} \frac{I_n}{J_n} $$

To find this limit, we need to evaluate both $I_n$ and $J_n$ and then find their ratio as $n \to \infty$.

Step 1: Find an antiderivative of $f(x)$

We have:

$$f(x) = \arctan\frac{1}{x} - \frac{1}{1+x^2}$$

Let's find $\int f(x) \, dx$ by integrating each term separately.

First term: $\int \arctan\frac{1}{x} \, dx$

Let $u = \arctan\frac{1}{x}$. We use integration by parts with:

$$u = \arctan\frac{1}{x}, \quad dv = dx$$

Then:

$$du = \frac{1}{1+\frac{1}{x^2}} \cdot \left(-\frac{1}{x^2}\right) dx = \frac{-1}{x^2+1} \, dx, \quad v = x$$

So:

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

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

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

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

Therefore:

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

Second term: $\int \frac{1}{1+x^2} \, dx$

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

Combining:

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

Step 2: Evaluate $I_n = \int_0^n f(x) \, dx$

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

At $x = n$:

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

At $x = 0$:

This requires careful analysis:

• $\lim_{x \to 0^+} x\arctan\frac{1}{x}$: As $x \to 0^+$, $\frac{1}{x} \to +\infty$, so $\arctan\frac{1}{x} \to \frac{\pi}{2}$ Using L'Hôpital's rule or noting that $x \cdot \frac{\pi}{2} \to 0$: $$ \lim_{x \to 0^+} x\arctan\frac{1}{x} = \lim_{x \to 0^+} \frac{\arctan\frac{1}{x}}{\frac{1}{x}} = \lim_{u \to \infty} \frac{\arctan u}{u} = 0 $$
• $\lim_{x \to 0^+} \frac{1}{2}\ln(x^2+1) = \frac{1}{2}\ln(1) = 0$
• $\lim_{x \to 0^+} \arctan x = \arctan 0 = 0$

Therefore, at $x = 0$: the value is $0 - 0 - 0 = 0$.

So:

$$I_n = n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) - \arctan n$$

Step 3: Evaluate $J_n = \int_{-n}^0 f(x) \, dx$

$$J_n = \left[x\arctan\frac{1}{x} + \frac{1}{2}\ln(x^2+1) - \arctan x\right]_{-n}^0$$

At $x = 0$: we found this is $0$.

At $x = -n$:

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

Using $\arctan(-u) = -\arctan(u)$:

$$= (-n)\left(-\arctan\frac{1}{n}\right) + \frac{1}{2}\ln(n^2+1) - (-\arctan n)$$

$$= n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) + \arctan n$$

Therefore:

$$J_n = 0 - \left[n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) + \arctan n\right]$$

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

Step 4: Find the ratio $\frac{I_n}{J_n}$

$$\frac{I_n}{J_n} = \frac{n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) - \arctan n}{-n\arctan\frac{1}{n} - \frac{1}{2}\ln(n^2+1) - \arctan n}$$

Let's denote:

$$A_n = n\arctan\frac{1}{n}, \quad B_n = \frac{1}{2}\ln(n^2+1), \quad C_n = \arctan n$$

Then:

$$\frac{I_n}{J_n} = \frac{A_n + B_n - C_n}{-A_n - B_n - C_n} = \frac{A_n + B_n - C_n}{-(A_n + B_n + C_n)}$$

$$= -\frac{A_n + B_n - C_n}{A_n + B_n + C_n}$$

Step 5: Analyze the limit as $n \to \infty$

As $n \to \infty$:

For $A_n = n\arctan\frac{1}{n}$:

Using the Taylor series $\arctan u = u - \frac{u^3}{3} + O(u^5)$ for small $u$:

$$\arctan\frac{1}{n} = \frac{1}{n} - \frac{1}{3n^3} + O\left(\frac{1}{n^5}\right)$$

$$A_n = n\arctan\frac{1}{n} = n\left(\frac{1}{n} - \frac{1}{3n^3} + O\left(\frac{1}{n^5}\right)\right) = 1 - \frac{1}{3n^2} + O\left(\frac{1}{n^4}\right)$$

So $\lim_{n \to \infty} A_n = 1$.

For $B_n = \frac{1}{2}\ln(n^2+1)$:

$$B_n = \frac{1}{2}\ln(n^2+1) = \frac{1}{2}\ln n^2\left(1+\frac{1}{n^2}\right) = \frac{1}{2}\left[\ln n^2 + \ln\left(1+\frac{1}{n^2}\right)\right]$$

$$= \ln n + \frac{1}{2}\ln\left(1+\frac{1}{n^2}\right) \to \infty$$

For $C_n = \arctan n$:

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

So as $n \to \infty$:

$$A_n \to 1, \quad B_n \to \infty, \quad C_n \to \frac{\pi}{2}$$

The dominant term in both numerator and denominator is $B_n = \ln n$.

$$\frac{I_n}{J_n} = -\frac{A_n + B_n - C_n}{A_n + B_n + C_n} = -\frac{1 + \ln n - \frac{\pi}{2} + o(1)}{1 + \ln n + \frac{\pi}{2} + o(1)}$$

Dividing numerator and denominator by $\ln n$:

$$= -\frac{\frac{1}{\ln n} + 1 - \frac{\pi}{2\ln n} + o\left(\frac{1}{\ln n}\right)}{\frac{1}{\ln n} + 1 + \frac{\pi}{2\ln n} + o\left(\frac{1}{\ln n}\right)}$$

As $n \to \infty$, $\frac{1}{\ln n} \to 0$:

$$\lim_{n \to \infty} \frac{I_n}{J_n} = -\frac{0 + 1 - 0}{0 + 1 + 0} = -\frac{1}{1} = -1$$

Final Answer

$$\boxed{\lim_{n \to \infty} \frac{I_n}{J_n} = -1}$$

Interpretation

The limit being $-1$ means that as $n \to \infty$, the integrals $I_n$ and $J_n$ have equal magnitudes but opposite signs.

More precisely:

$$I_n \approx -J_n \quad \text{for large } n$$

This makes sense because:

• Both $I_n$ and $J_n$ grow to infinity (due to the $\ln n$ term)
• The asymmetry between the two integrals comes from the $\arctan n$ term, which contributes $-\frac{\pi}{2}$ to $I_n$ and $-\frac{\pi}{2}$ to $J_n$, canceling out in the ratio
• The dominant behavior is captured by $B_n = \frac{1}{2}\ln(n^2+1)$, which appears with opposite signs in the two integrals

Summary

1. We found the antiderivative: $$ \int f(x) \, dx = x\arctan\frac{1}{x} + \frac{1}{2}\ln(x^2+1) - \arctan x + C $$
2. We evaluated: $$ I_n = n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) - \arctan n $$ $$ J_n = -n\arctan\frac{1}{n} - \frac{1}{2}\ln(n^2+1) - \arctan n $$
3. We formed the ratio: $$ \frac{I_n}{J_n} = -\frac{n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) - \arctan n}{n\arctan\frac{1}{n} + \frac{1}{2}\ln(n^2+1) + \arctan n} $$
4. The dominant term is $\frac{1}{2}\ln(n^2+1) \sim \ln n$, leading to: $$ \lim_{n \to \infty} \frac{I_n}{J_n} = -1 $$

Key lessons:

• When evaluating limits of ratios, identify the dominant terms
• Integration by parts is essential for integrating $\arctan\frac{1}{x}$
• Careful analysis of limits at boundary points is crucial (especially at $x = 0$)
• Asymptotic analysis (Taylor series) helps identify dominant behavior as $n \to \infty$