How Do I Show This Real Integral Vanishes Only If One Of Its Arguments Is Zero?

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Introduction

In real analysis, integrals play a crucial role in understanding various mathematical concepts and their applications. The given integral, I(α,β)I(\alpha,\beta), is a complex expression involving the floor function, trigonometric functions, and exponential functions. In this article, we will delve into the properties of this integral and explore the conditions under which it vanishes.

Defining the Integral

The integral I(α,β)I(\alpha,\beta) is defined as:

I(α,β)=∫1∞(x−[x]−1)x−3/2(xα+x−α)sin⁡(βln⁡(x)) dx,I(\alpha,\beta) = \int_1^\infty (x-[x]-1)x^{-3/2}(x^\alpha+x^{-\alpha})\sin(\beta \ln(x))\,dx,

where [x][x] is the floor function, and the parameters α\alpha and β\beta are restricted to the following ranges:

−1/2<α<1/2,β∈R.-1/2 < \alpha < 1/2, \quad \beta \in \mathbb{R}.

Breaking Down the Integral

To understand the behavior of the integral, let's break it down into its constituent parts. The integral can be rewritten as:

I(α,β)=∫1∞(x−[x]−1)x−3/2(xα+x−α)sin⁡(βln⁡(x)) dx=∫1∞(x−[x]−1)x−3/2xαsin⁡(βln⁡(x)) dx+∫1∞(x−[x]−1)x−3/2x−αsin⁡(βln⁡(x)) dxI(\alpha,\beta) = \int_1^\infty (x-[x]-1)x^{-3/2}(x^\alpha+x^{-\alpha})\sin(\beta \ln(x))\,dx = \int_1^\infty (x-[x]-1)x^{-3/2}x^\alpha\sin(\beta \ln(x))\,dx + \int_1^\infty (x-[x]-1)x^{-3/2}x^{-\alpha}\sin(\beta \ln(x))\,dx

The first term on the right-hand side can be further simplified by using the fact that x−[x]−1x-[x]-1 is equal to 00 for all integer values of xx. This means that the first term is equal to 00 for all integer values of xx. Therefore, we can rewrite the integral as:

I(α,β)=∫1∞(x−[x]−1)x−3/2x−αsin⁡(βln⁡(x)) dxI(\alpha,\beta) = \int_1^\infty (x-[x]-1)x^{-3/2}x^{-\alpha}\sin(\beta \ln(x))\,dx

Understanding the Floor Function

The floor function, denoted by [x][x], is a mathematical function that returns the greatest integer less than or equal to xx. This function plays a crucial role in the given integral, as it affects the behavior of the integrand. To understand the behavior of the floor function, let's consider the following:

x−[x]−1={0,if x is an integer,1,if x is not an integer.x-[x]-1 = \begin{cases} 0, & \text{if } x \text{ is an integer}, \\ 1, & \text{if } x \text{ is not an integer}. \end{cases}

This means that the floor function is equal to 00 for all integer values of xx, and equal to 11 for all non-integer values of xx.

Analyzing the Integral

To analyze the integral, let's consider the following:

I(α,β)=∫1∞(x−[x]−1)x−3/2x−αsin⁡(βln⁡(x)) dxI(\alpha,\beta) = \int_1^\infty (x-[x]-1)x^{-3/2}x^{-\alpha}\sin(\beta \ln(x))\,dx

Using the fact that x−[x]−1x-[x]-1 is equal to 00 for all integer values of xx, we can rewrite the integral as:

I(α,β)=∫1∞x−3/2x−αsin⁡(βln⁡(x)) dxI(\alpha,\beta) = \int_1^\infty x^{-3/2}x^{-\alpha}\sin(\beta \ln(x))\,dx

This integral can be further simplified by using the following substitution:

u=ln⁡(x)u = \ln(x)

This substitution leads to the following:

I(α,β)=∫0∞e−3u/2e−αusin⁡(βu) duI(\alpha,\beta) = \int_0^\infty e^{-3u/2}e^{-\alpha u}\sin(\beta u)\,du

Understanding the Exponential Function

The exponential function, denoted by exe^x, is a mathematical function that is defined as:

ex=∑n=0∞xnn!e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

This function plays a crucial role in the given integral, as it affects the behavior of the integrand. To understand the behavior of the exponential function, let's consider the following:

e−3u/2e−αu=∑n=0∞(−3u/2)nn!∑m=0∞(−αu)mm!e^{-3u/2}e^{-\alpha u} = \sum_{n=0}^\infty \frac{(-3u/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha u)^m}{m!}

This means that the exponential function is a sum of an infinite number of terms, each of which is a product of two exponential functions.

Analyzing the Integral (continued)

To analyze the integral, let's consider the following:

I(α,β)=∫0∞e−3u/2e−αusin⁡(βu) duI(\alpha,\beta) = \int_0^\infty e^{-3u/2}e^{-\alpha u}\sin(\beta u)\,du

Using the fact that the exponential function is a sum of an infinite number of terms, we can rewrite the integral as:

I(α,β)=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∫0∞sin⁡(βu)un+m duI(\alpha,\beta) = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \int_0^\infty \sin(\beta u)u^{n+m}\,du

This integral can be further simplified by using the following substitution:

v=u2v = u^2

This substitution leads to the following:

I(α,β)=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∫0∞sin⁡(βu)un+m du=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∫0∞sin⁡(βu)un+m du=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∫0∞sin⁡(βu)un+m duI(\alpha,\beta) = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \int_0^\infty \sin(\beta u)u^{n+m}\,du = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \int_0^\infty \sin(\beta u)u^{n+m}\,du = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \int_0^\infty \sin(\beta u)u^{n+m}\,du

Understanding the Sine Function

The sine function, denoted by sin⁡(x)\sin(x), is a mathematical function that is defined as:

sin⁡(x)=∑n=0∞(−1)nx2n+1(2n+1)!\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}

This function plays a crucial role in the given integral, as it affects the behavior of the integrand. To understand the behavior of the sine function, let's consider the following:

sin⁡(βu)=∑n=0∞(−1)n(βu)2n+1(2n+1)!\sin(\beta u) = \sum_{n=0}^\infty \frac{(-1)^n (\beta u)^{2n+1}}{(2n+1)!}

This means that the sine function is a sum of an infinite number of terms, each of which is a product of two exponential functions.

Analyzing the Integral (continued)

To analyze the integral, let's consider the following:

I(α,β)=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∫0∞sin⁡(βu)un+m duI(\alpha,\beta) = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \int_0^\infty \sin(\beta u)u^{n+m}\,du

Using the fact that the sine function is a sum of an infinite number of terms, we can rewrite the integral as:

I(α,β)=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∑k=0∞(−1)k(β)2k+1(2k+1)!∫0∞un+m+2k+1 duI(\alpha,\beta) = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \sum_{k=0}^\infty \frac{(-1)^k (\beta)^{2k+1}}{(2k+1)!} \int_0^\infty u^{n+m+2k+1}\,du

This integral can be further simplified by using the following substitution:

w=un+m+2k+2w = u^{n+m+2k+2}

This substitution leads to the following:

I(α,β)=∑n=0∞(−3/2)nn!∑m=0∞(−α)mm!∑k=0∞(−1)k(β)2k+1(2k+1)!∫0∞w(n+m+2k+1)/2 dwI(\alpha,\beta) = \sum_{n=0}^\infty \frac{(-3/2)^n}{n!} \sum_{m=0}^\infty \frac{(-\alpha)^m}{m!} \sum_{k=0}^\infty \frac{(-1)^k (\beta)^{2k+1}}{(2k+1)!} \int_0^\infty w^{(n+m+2k+1)/2}\,dw

Understanding the Gamma Function

The gamma function, denoted by Γ(x)\Gamma(x), is a mathematical function that is defined as:

\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}\,dt<br/> **Q&A: Understanding the Real Integral and its Vanishing Conditions** ==================================================================== **Q: What is the given integral, and what are its parameters?** --------------------------------------------------------- A: The given integral is $I(\alpha,\beta)$, which is defined as:

I(\alpha,\beta) = \int_1^\infty (x-[x]-1)x{-3/2}(x\alpha+x^{-\alpha})\sin(\beta \ln(x)),dx,

where $[x]$ is the floor function, and the parameters $\alpha$ and $\beta$ are restricted to the following ranges:

-1/2 < \alpha < 1/2, \quad \beta \in \mathbb{R}.

**Q: What is the significance of the floor function in the integral?** ---------------------------------------------------------------- A: The floor function, denoted by $[x]$, is a mathematical function that returns the greatest integer less than or equal to $x$. In the given integral, the floor function affects the behavior of the integrand, particularly when $x$ is an integer. **Q: How does the integral behave when $x$ is an integer?** ------------------------------------------------------ A: When $x$ is an integer, the floor function $[x]$ is equal to $x$. Therefore, the term $(x-[x]-1)$ is equal to $0$, and the integral reduces to:

I(\alpha,\beta) = \int_1^\infty x{-3/2}x{-\alpha}\sin(\beta \ln(x)),dx

**Q: What is the significance of the exponential function in the integral?** ------------------------------------------------------------------- A: The exponential function, denoted by $e^x$, is a mathematical function that is defined as:

e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

In the given integral, the exponential function affects the behavior of the integrand, particularly when $x$ is large. **Q: How does the integral behave when $x$ is large?** --------------------------------------------------- A: When $x$ is large, the term $x^{-3/2}$ dominates the integrand, and the integral reduces to:

I(\alpha,\beta) = \int_1^\infty x^{-3/2}\sin(\beta \ln(x)),dx

**Q: What is the significance of the sine function in the integral?** ---------------------------------------------------------------- A: The sine function, denoted by $\sin(x)$, is a mathematical function that is defined as:

\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}

In the given integral, the sine function affects the behavior of the integrand, particularly when $\beta$ is large. **Q: How does the integral behave when $\beta$ is large?** ------------------------------------------------------ A: When $\beta$ is large, the term $\sin(\beta \ln(x))$ dominates the integrand, and the integral reduces to:

I(\alpha,\beta) = \int_1^\infty x^{-3/2}\sin(\beta \ln(x)),dx

**Q: What are the conditions under which the integral vanishes?** --------------------------------------------------------- A: The integral vanishes if one of the following conditions is met: * $\alpha = 0$ * $\beta = 0$ * $x = 0$ **Q: Why does the integral vanish under these conditions?** ------------------------------------------------------ A: The integral vanishes under these conditions because the integrand becomes zero. Specifically: * When $\alpha = 0$, the term $x^{-\alpha}$ becomes $1$, and the integrand reduces to $x^{-3/2}\sin(\beta \ln(x))$, which is zero when $\beta = 0$. * When $\beta = 0$, the term $\sin(\beta \ln(x))$ becomes $0$, and the integrand reduces to $x^{-3/2}$, which is zero when $x = 0$. * When $x = 0$, the integrand becomes zero because the term $x^{-3/2}$ becomes infinite. **Q: What are the implications of the integral vanishing under these conditions?** ------------------------------------------------------------------------- A: The integral vanishing under these conditions has significant implications for the behavior of the function $I(\alpha,\beta)$. Specifically: * When $\alpha = 0$, the function $I(\alpha,\beta)$ becomes zero for all values of $\beta$. * When $\beta = 0$, the function $I(\alpha,\beta)$ becomes zero for all values of $\alpha$. * When $x = 0$, the function $I(\alpha,\beta)$ becomes zero for all values of $\alpha$ and $\beta$. **Q: What are the applications of the integral vanishing under these conditions?** ------------------------------------------------------------------------- A: The integral vanishing under these conditions has significant applications in various fields, including: * **Mathematics**: The integral vanishing under these conditions has implications for the behavior of functions and their derivatives. * **Physics**: The integral vanishing under these conditions has implications for the behavior of physical systems and their properties. * **Engineering**: The integral vanishing under these conditions has implications for the design and analysis of engineering systems. **Conclusion** ---------- In conclusion, the integral $I(\alpha,\beta)$ has significant implications for the behavior of functions and their derivatives. The integral vanishes under certain conditions, including when $\alpha = 0$, $\beta = 0$, or $x = 0$. These conditions have significant implications for the behavior of the function $I(\alpha,\beta)$ and its applications in various fields.