Find: lim as t to infinity (integral from 1 to t (arctan

Find: $$ \lim_{t \to \infty} \frac{\int_1^t \left(\arctan\frac{x+1}{x-1} - \frac{\pi}{4}\right) dx}{\ln t} $$

To find this limit, we need to evaluate the integral and then analyze its asymptotic behavior.

Step 1: Simplify the integrand

First, let's simplify $\arctan\frac{x+1}{x-1}$ using the arctangent subtraction formula.

Recall that:

$$\arctan a - \arctan b = \arctan\frac{a-b}{1+ab}$$

We can write $\frac{x+1}{x-1}$ in a useful form. Note that:

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

Using the arctangent addition formula:

$$\arctan(u+v) = \arctan u + \arctan v - \arctan(uv) \quad \text{(if certain conditions hold)}$$

Actually, let's use a different approach. We have:

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

Using the identity $\arctan\frac{a+b}{1-ab} = \arctan a + \arctan b$ (when $ab < 1$), or more directly:

Let's write:

$$\frac{x+1}{x-1} = \frac{x-1+2}{x-1} = 1 + \frac{2}{x-1}$$

For $x > 1$, we can use:

$$\arctan\frac{x+1}{x-1} - \frac{\pi}{4} = \arctan\frac{x+1}{x-1} - \arctan 1$$

Using the arctangent subtraction formula:

$$\arctan a - \arctan b = \arctan\frac{a-b}{1+ab}$$

With $a = \frac{x+1}{x-1}$ and $b = 1$:

$$\arctan\frac{x+1}{x-1} - \arctan 1 = \arctan\frac{\frac{x+1}{x-1} - 1}{1 + \frac{x+1}{x-1}}$$

Simplifying the numerator:

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

Simplifying the denominator:

$$1 + \frac{x+1}{x-1} = \frac{x-1 + x+1}{x-1} = \frac{2x}{x-1}$$

Therefore:

$$\arctan\frac{x+1}{x-1} - \frac{\pi}{4} = \arctan\frac{\frac{2}{x-1}}{\frac{2x}{x-1}} = \arctan\frac{2}{2x} = \arctan\frac{1}{x}$$

Great! The integrand simplifies to:

$$\arctan\frac{x+1}{x-1} - \frac{\pi}{4} = \arctan\frac{1}{x}$$

Step 2: Evaluate the integral

Now we need to find:

$$\int_1^t \arctan\frac{1}{x} \, dx$$

Integration by parts:

Let $u = \arctan\frac{1}{x}$ and $dv = dx$.

Then:

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

$$v = x$$

Using integration by parts:

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

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

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:

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

Step 3: Evaluate the definite integral

$$\int_1^t \arctan\frac{1}{x} \, dx = \left[x\arctan\frac{1}{x} + \frac{1}{2}\ln(x^2+1)\right]_1^t$$

At $x = t$:

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

At $x = 1$:

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

Therefore:

$$\int_1^t \arctan\frac{1}{x} \, dx = t\arctan\frac{1}{t} + \frac{1}{2}\ln(t^2+1) - \frac{\pi}{4} - \frac{1}{2}\ln 2$$

Step 4: Analyze the asymptotic behavior as $t \to \infty$

We have:

$$\int_1^t \arctan\frac{1}{x} \, dx = t\arctan\frac{1}{t} + \frac{1}{2}\ln(t^2+1) - \frac{\pi}{4} - \frac{1}{2}\ln 2$$

As $t \to \infty$:

For $t\arctan\frac{1}{t}$:

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

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

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

So $\lim_{t \to \infty} t\arctan\frac{1}{t} = 1$.

For $\frac{1}{2}\ln(t^2+1)$:

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

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

Using $\ln(1+u) = u + O(u^2)$ for small $u$:

$$= \ln t + \frac{1}{2} \cdot \frac{1}{t^2} + O\left(\frac{1}{t^4}\right) = \ln t + O\left(\frac{1}{t^2}\right)$$

Therefore, as $t \to \infty$:

$$\int_1^t \arctan\frac{1}{x} \, dx = 1 + \ln t - \frac{\pi}{4} - \frac{1}{2}\ln 2 + O\left(\frac{1}{t^2}\right)$$

$$= \ln t + \left(1 - \frac{\pi}{4} - \frac{1}{2}\ln 2\right) + O\left(\frac{1}{t^2}\right)$$

Step 5: Calculate the limit

$$\lim_{t \to \infty} \frac{\int_1^t \arctan\frac{1}{x} \, dx}{\ln t} = \lim_{t \to \infty} \frac{\ln t + \left(1 - \frac{\pi}{4} - \frac{1}{2}\ln 2\right) + O\left(\frac{1}{t^2}\right)}{\ln t}$$

$$= \lim_{t \to \infty} \left[1 + \frac{1 - \frac{\pi}{4} - \frac{1}{2}\ln 2}{\ln t} + \frac{O\left(\frac{1}{t^2}\right)}{\ln t}\right]$$

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

$$\lim_{t \to \infty} \frac{\int_1^t \arctan\frac{1}{x} \, dx}{\ln t} = 1 + 0 + 0 = 1$$

Final Answer

$$\boxed{\lim_{t \to \infty} \frac{\int_1^t \left(\arctan\frac{x+1}{x-1} - \frac{\pi}{4}\right) dx}{\ln t} = 1}$$

Interpretation

The result means that the integral grows like $\ln t$ as $t \to \infty$. More precisely:

$$\int_1^t \left(\arctan\frac{x+1}{x-1} - \frac{\pi}{4}\right) dx \sim \ln t$$

This makes sense because:

• The integrand $\arctan\frac{1}{x}$ behaves like $\frac{1}{x}$ for large $x$
• We know $\int_1^t \frac{1}{x} \, dx = \ln t$
• The integral of $\arctan\frac{1}{x}$ has the same logarithmic growth, plus lower-order terms

Summary

1. We simplified the integrand using the arctangent subtraction formula: $$ \arctan\frac{x+1}{x-1} - \frac{\pi}{4} = \arctan\frac{1}{x} $$
2. We evaluated the integral using integration by parts: $$ \int_1^t \arctan\frac{1}{x} \, dx = t\arctan\frac{1}{t} + \frac{1}{2}\ln(t^2+1) - \frac{\pi}{4} - \frac{1}{2}\ln 2 $$
3. We performed asymptotic analysis: $$ \int_1^t \arctan\frac{1}{x} \, dx = \ln t + C + O\left(\frac{1}{t^2}\right) $$ where $C = 1 - \frac{\pi}{4} - \frac{1}{2}\ln 2$ is a constant.
4. The dominant term is $\ln t$, giving the limit: $$ \lim_{t \to \infty} \frac{\int_1^t \arctan\frac{1}{x} \, dx}{\ln t} = 1 $$

Key lessons:

• Arctangent subtraction formulas can dramatically simplify expressions
• Integration by parts is essential for integrals involving inverse trigonometric functions
• Asymptotic analysis using Taylor series reveals the dominant behavior
• When finding limits of ratios, identify and compare the dominant terms
• The integral of $\arctan\frac{1}{x}$ grows logarithmically, similar to $\int \frac{1}{x} \, dx$