Halla el límite de la integral

$$ \lim_{x \to +\infty} \frac{1}{x} \int_1^x \frac{1}{t} \tan\left(\frac{\pi t}{2t + 1}\right) \, dt, $$

RESPUESTA

To evaluate the limit

$$ \lim_{x \to +\infty} \frac{1}{x} \int_1^x \frac{1}{t} \tan\left(\frac{\pi t}{2t + 1}\right) \, dt, $$

we begin by analyzing the behavior of the integrand as $ t \to \infty $.

# **Step 1: Simplify the Argument of the Tangent**

The argument of the tangent function is:

$$ \frac{\pi t}{2t + 1}.$$

Dividing numerator and denominator by $ t $, we get:

$$ \frac{\pi t}{2t + 1} = \frac{\pi}{2 + \frac{1}{t}}. $$

As $ t \to \infty $, this tends to:

$$ \frac{\pi}{2}. $$

So the argument of the tangent function approaches $ \frac{\pi}{2} $, where the tangent function diverges. However, we can approximate the behavior of the tangent near $ \frac{\pi}{2} $ using the identity:

$$ \tan\left(\frac{\pi}{2} - \varepsilon\right) = \cot(\varepsilon) \approx \frac{1}{\varepsilon}, \quad \text{for small } \varepsilon. $$

### **Step 2: Approximate the Tangent Function**

Let us define:

$$ \varepsilon = \frac{\pi}{2} - \frac{\pi t}{2t + 1} = \frac{\pi}{2(2t + 1)}. $$

Then, $$ \tan\left(\frac{\pi t}{2t + 1}\right) = \tan\left(\frac{\pi}{2} - \varepsilon\right) \approx \frac{1}{\varepsilon} = \frac{2(2t + 1)}{\pi}. $$

Thus, the integrand becomes:

$$ \frac{1}{t} \tan\left(\frac{\pi t}{2t + 1}\right) \approx \frac{1}{t} \cdot \frac{2(2t + 1)}{\pi} = \frac{4t + 2}{\pi t} = \frac{4}{\pi} + \frac{2}{\pi t}. $$

As $ t \to \infty $, the second term becomes negligible, and the integrand approaches:

$$ \frac{4}{\pi}.$$

# **Step 3: Estimate the Integral**

Since the integrand approaches a constant $ \frac{4}{\pi} $ as $ t \to \infty $, the integral from 1 to $ x $ behaves approximately like:

$$ \int_1^x \frac{4}{\pi} \, dt = \frac{4}{\pi}(x - 1). $$

Therefore, the original expression becomes:

$$ \frac{1}{x} \int_1^x \frac{1}{t} \tan\left(\frac{\pi t}{2t + 1}\right) dt \approx \frac{1}{x} \cdot \frac{4}{\pi}(x - 1) = \frac{4}{\pi} \cdot \left(1 - \frac{1}{x}\right). $$

As $ x \to \infty $, the term $ \frac{1}{x} \to 0 $, and the expression converges to:

$$ \frac{4}{\pi}. $$

## **Step 4: Confirm with L’Hôpital’s Rule**

We can also apply L’Hôpital’s Rule to the limit:

$$ \lim_{x \to \infty} \frac{\int_1^x \frac{1}{t} \tan\left(\frac{\pi t}{2t + 1}\right) dt}{x}. $$

This is an indeterminate form $ \frac{\infty}{\infty} $, so we differentiate numerator and denominator:

- The derivative of the numerator is the integrand evaluated at $ x $:

$$ \frac{1}{x} \tan\left(\frac{\pi x}{2x + 1}\right), $$

- The derivative of the denominator is 1.

So the limit becomes: $$ \lim_{x \to \infty} \frac{1}{x} \tan\left(\frac{\pi x}{2x + 1}\right). $$

As shown earlier, this expression approaches: $$ \frac{4}{\pi}. $$

# **Conclusion**

Both the direct approximation and L’Hôpital’s Rule confirm that the limit is:

$$ \boxed{\dfrac{4}{\pi}} $$

otra respuesta

Para resolver el límite:

$\lim_{x \to + \infty} \frac{1}{x} \int_{1}^{x} \frac{1}{t} \tan (\frac{π t}{2 t + 1}) d t,$

analizamos el comportamiento del integrando cuando $t \to \infty$ .

Paso 1: Simplificar el argumento de la tangente

El argumento de $\tan$ es $\frac{π t}{2 t + 1}$ . Dividiendo numerador y denominador por $t$ :

$\frac{π t}{2 t + 1} = \frac{π}{2 + \frac{1}{t}} \approx \frac{π}{2} - \frac{π}{4 t} para t \to \infty .$

Paso 2: Aproximar $\tan (\frac{π t}{2 t + 1})$

Usando la aproximación $\tan (\frac{π}{2} - ϵ) \approx \cot (ϵ)$ para $ϵ \to 0$ :

$\tan (\frac{π}{2} - \frac{π}{4 t}) \approx \cot (\frac{π}{4 t}) \approx \frac{4 t}{π} .$

Paso 3: Comportamiento del integrando

El integrando se aproxima como:

$\frac{1}{t} \tan (\frac{π t}{2 t + 1}) \approx \frac{1}{t} \cdot \frac{4 t}{π} = \frac{4}{π} .$

Paso 4: Evaluar la integral

Integramos la aproximación constante:

$\int_{1}^{x} \frac{4}{π} d t = \frac{4}{π} (x - 1) .$

Paso 5: Calcular el límite

Dividimos por $x$ y tomamos el límite:

$\lim_{x \to + \infty} \frac{1}{x} \cdot \frac{4}{π} (x - 1) = \frac{4}{π} .$

Conclusión

El límite es:

$\frac{4}{π}$

Otra respuesta

0

To solve the limit problem:

\lim_{x \to + \infty} \frac{1}{x} \int_{x}^{11 x} \frac{\tan (π t)}{2 t + 1} d t

we analyze the integral and its behavior as $x$ approaches infinity. The integrand $\tan (π t) / (2 t + 1)$ is oscillatory with period 1 and has singularities at $t = n + 1 / 2$ for integers $n$ . However, the integral can be interpreted in the principal value sense to handle these singularities.

Key steps:

Periodicity and Oscillatory Nature: The function $\tan (π t)$ is periodic with period 1 and oscillates between $- \infty$ and $+ \infty$ with singularities at $t = n + 1 / 2$ . The integrand $\tan (π t) / (2 t + 1)$ decays as $1 / t$ due to the denominator.
Principal Value Integral: The integral over each interval around the singularities is considered in the principal value sense, leading to cancellation of the divergent parts.
Integration by Parts: Using integration by parts, we show that the boundary terms and the remaining integral term both tend to zero when divided by $x$ .
Riemann-Lebesgue Lemma: The oscillatory nature of $\tan (π t)$ and the decay of $1 / (2 t + 1)$ imply that the integral averages out to zero over large intervals.

After detailed analysis, we conclude that the integral $\int_{x}^{11 x} \frac{\tan (π t)}{2 t + 1} d t$ grows slower than $x$ , leading to the limit:

\lim_{x \to + \infty} \frac{1}{x} \int_{x}^{11 x} \frac{\tan (π t)}{2 t + 1} d t = 0

0

RESPUESTA

To evaluate the limit

$x \to + \infty lim \frac{1}{x} \int_{1}^{x} \frac{1}{t} tan (\frac{π t}{2 t + 1}) d t,$

we begin by analyzing the behavior of the integrand as $t \to \infty$ .

Step 1: Simplify the Argument of the Tangent

The argument of the tangent function is:

$\frac{π t}{2 t + 1} .$

Dividing numerator and denominator by $t$ , we get:

$\frac{π t}{2 t + 1} = \frac{π}{2 + \frac{1}{t}} .$

As $t \to \infty$ , this tends to:

$\frac{π}{2} .$

So the argument of the tangent function approaches $\frac{π}{2}$ , where the tangent function diverges. However, we can approximate the behavior of the tangent near $\frac{π}{2}$ using the identity:

$tan (\frac{π}{2} - ε) = cot (ε) \approx \frac{1}{ε}, for small ε .$

Step 2: Approximate the Tangent Function

Let us define:

$ε = \frac{π}{2} - \frac{π t}{2 t + 1} = \frac{π}{2 ( 2 t + 1 )} .$

Then,

$tan (\frac{π t}{2 t + 1}) = tan (\frac{π}{2} - ε) \approx \frac{1}{ε} = \frac{2 ( 2 t + 1 )}{π} .$

Thus, the integrand becomes:

$\frac{1}{t} tan (\frac{π t}{2 t + 1}) \approx \frac{1}{t} \cdot \frac{2 ( 2 t + 1 )}{π} = \frac{4 t + 2}{π t} = \frac{4}{π} + \frac{2}{π t} .$

As $t \to \infty$ , the second term becomes negligible, and the integrand approaches:

$\frac{4}{π} .$

Step 3: Estimate the Integral

Since the integrand approaches a constant $\frac{4}{π}$ as $t \to \infty$ , the integral from 1 to $x$ behaves approximately like:

$\int_{1}^{x} \frac{4}{π} d t = \frac{4}{π} (x - 1) .$

Therefore, the original expression becomes:

$\frac{1}{x} \int_{1}^{x} \frac{1}{t} tan (\frac{π t}{2 t + 1}) d t \approx \frac{1}{x} \cdot \frac{4}{π} (x - 1) = \frac{4}{π} \cdot (1 - \frac{1}{x}) .$

As $x \to \infty$ , the term $\frac{1}{x} \to 0$ , and the expression converges to:

$\frac{4}{π} .$

Step 4: Confirm with L’Hôpital’s Rule

We can also apply L’Hôpital’s Rule to the limit:

$x \to \infty lim \frac{\int _{1}^{x} \frac{1}{t} tan ( \frac{π t}{2 t + 1} ) d t}{x} .$

This is an indeterminate form $\frac{\infty}{\infty}$ , so we differentiate numerator and denominator:

The derivative of the numerator is the integrand evaluated at $x$ :
$\frac{1}{x} tan (\frac{π x}{2 x + 1}),$
The derivative of the denominator is 1.

So the limit becomes:

$x \to \infty lim \frac{1}{x} tan (\frac{π x}{2 x + 1}) .$

As shown earlier, this expression approaches:

$\frac{4}{π} .$

Conclusion

Both the direct approximation and L’Hôpital’s Rule confirm that the limit is:

$\frac{4}{π}$

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