How To Compute $\int_0^\infty \frac{\sin T}{t^{s+1}} Dt $?

Introduction

The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is a complex and challenging problem that has been studied extensively in the field of complex analysis. The integral is defined as $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt$, where the real part of the complex number $s$ is negative and greater than $-1$. In this article, we will explore the methods and techniques used to compute this integral.

Background and Motivation

The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is a special case of the more general integral of $\frac{\sin t}{t^s}$ from $0$ to $\infty$. This integral has been studied in various contexts, including complex analysis, trigonometry, and mathematical physics. The integral is of particular interest because it appears in the calculation of the gamma function, which is a fundamental function in mathematics.

The Gamma Function

The gamma function is defined as $\Gamma(s) = \int_0^\infty t^{s-1} e^{-t} \;\text dt$. The gamma function is an extension of the factorial function to real and complex numbers. It is defined for all complex numbers except for non-positive integers. The gamma function has many important properties, including the reflection formula, which states that $\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin \pi s}$.

The Integral of $\frac{\sin t}{t^{s+1}}$

To compute the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$, we can use the following approach:

1. Substitution: Let $u = t^s$. Then, $du = st^{s-1} dt$.
2. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt = \frac{1}{s} \displaystyle\int_0^\infty \frac{\sin t}{t^s} \;\text dt$.
3. Change of Variables: Let $v = t^s$. Then, $dv = st^{s-1} dt$.
4. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin t}{t^s} \;\text dt = \frac{1}{s} \displaystyle\int_0^\infty \frac{\sin v^{1/s}}{v} \;\text dv$.

Computing the Integral

To compute the integral of $\frac{\sin v^{1/s}}{v}$ from $0$ to $\infty$, we can use the following approach:

1. Substitution: Let $w = v^{1/s}$. Then, $dw = \frac{1}{s} v^{(1/s)-1} dv$.
2. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin v^{1/s}}{v} \;\text dv = \displaystyle\int_0^\infty \frac{\sin w}{w} \;\text dw$.
3. Evaluation: The integral of $\frac{\sin w}{w}$ from $0$ to $\infty$ is a well-known result in mathematics, which is equal to $\frac{\pi}{2}$.

Conclusion

In conclusion, the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ can be computed using the following approach:

1. Substitution: Let $u = t^s$. Then, $du = st^{s-1} dt$.
2. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt = \frac{1}{s} \displaystyle\int_0^\infty \frac{\sin t}{t^s} \;\text dt$.
3. Change of Variables: Let $v = t^s$. Then, $dv = st^{s-1} dt$.
4. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin t}{t^s} \;\text dt = \frac{1}{s} \displaystyle\int_0^\infty \frac{\sin v^{1/s}}{v} \;\text dv$.
5. Substitution: Let $w = v^{1/s}$. Then, $dw = \frac{1}{s} v^{(1/s)-1} dv$.
6. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin v^{1/s}}{v} \;\text dv = \displaystyle\int_0^\infty \frac{\sin w}{w} \;\text dw$.
7. Evaluation: The integral of $\frac{\sin w}{w}$ from $0$ to $\infty$ is a well-known result in mathematics, which is equal to $\frac{\pi}{2}$.

Q: What is the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$?

A: The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is a complex and challenging problem that has been studied extensively in the field of complex analysis. The integral is defined as $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt$, where the real part of the complex number $s$ is negative and greater than $-1$.

Q: How do I compute the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$?

A: To compute the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$, you can use the following approach:

1. Substitution: Let $u = t^s$. Then, $du = st^{s-1} dt$.
2. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt = \frac{1}{s} \displaystyle\int_0^\infty \frac{\sin t}{t^s} \;\text dt$.
3. Change of Variables: Let $v = t^s$. Then, $dv = st^{s-1} dt$.
4. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin t}{t^s} \;\text dt = \frac{1}{s} \displaystyle\int_0^\infty \frac{\sin v^{1/s}}{v} \;\text dv$.
5. Substitution: Let $w = v^{1/s}$. Then, $dw = \frac{1}{s} v^{(1/s)-1} dv$.
6. Integration: The integral becomes $\displaystyle\int_0^\infty \frac{\sin v^{1/s}}{v} \;\text dv = \displaystyle\int_0^\infty \frac{\sin w}{w} \;\text dw$.
7. Evaluation: The integral of $\frac{\sin w}{w}$ from $0$ to $\infty$ is a well-known result in mathematics, which is equal to $\frac{\pi}{2}$.

Q: What is the final answer to the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$?

A: The final answer to the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is $\boxed{\frac{\pi}{2s}}$.

Q: What are some common applications of the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$?

A: The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ has many important applications in mathematics and physics. Some common applications include:

• Complex Analysis: The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is used to study the properties of complex functions and their behavior in the complex plane.
• Trigonometry: The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is used to study the properties of trigonometric functions and their behavior in the complex plane.
• Mathematical Physics: The integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ is used to study the properties of physical systems and their behavior in the complex plane.

Q: What are some common mistakes to avoid when computing the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$?

A: Some common mistakes to avoid when computing the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ include:

• Incorrect Substitution: Failing to use the correct substitution or using the wrong substitution can lead to incorrect results.
• Incorrect Integration: Failing to integrate the correct function or integrating the wrong function can lead to incorrect results.
• Incorrect Evaluation: Failing to evaluate the integral correctly or evaluating the integral incorrectly can lead to incorrect results.

Q: How can I practice computing the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$?

A: To practice computing the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$, you can try the following:

• Practice Problems: Try solving practice problems that involve computing the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$.
• Real-World Applications: Try applying the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$ to real-world problems in mathematics and physics.
• Online Resources: Try using online resources such as calculators and software to practice computing the integral of $\frac{\sin t}{t^{s+1}}$ from $0$ to $\infty$.