Method of Integration by Parts

Oct 13, 2025 by pynkocean.com 130 views

Does the Integral Converge? A Comprehensive Analysis of the Given Improper Integral

In this article, we will delve into the world of improper integrals and explore the convergence of a specific integral. The given integral is $ \int_0^\infty \exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx$, where $m>0$ . Our goal is to determine whether this integral converges for the given value of $m$ .

The integral in question is an improper integral, which means that it has an infinite limit of integration. This type of integral is often used to model real-world phenomena, such as the behavior of physical systems over an infinite time period. The given integral can be written as:

\int_0^\infty \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

To analyze the convergence of this integral, we need to understand the behavior of the integrand as $x$ approaches infinity.

Let's examine the integrand more closely:

\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\}

As $x$ approaches infinity, the term $\frac{m}{\sqrt{2}x}$ approaches zero. This is because the denominator $x$ becomes very large, causing the fraction to approach zero. Therefore, as $x$ approaches infinity, the integrand approaches:

\exp\left\{-\left[\frac{x}{\sqrt{2}}\right]^2\right\}

This is a Gaussian function, which decays exponentially as $x$ approaches infinity. This suggests that the integral may converge.

To determine whether the integral converges, we need to evaluate the integral over the infinite interval. One way to do this is to use the method of integration by parts.

Method of Integration by Parts

Let's integrate the integrand by parts:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

We can choose $u = \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\}$ and $dv = dx$ . Then, we have:

du = -2\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

v = x

Now, we can integrate by parts:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx = x\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} - \int x\left(-2\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\}\right) dx

Evaluating the integral on the right-hand side, we get:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx = x\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\int \left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

Now, we can integrate by parts again:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx = x\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} - 2\sqrt{2}\int \left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

Evaluating the integral on the right-hand side, we get:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx = x\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} - 2\sqrt{2}\int \left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

Now, we can integrate by parts again:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx = x\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} - 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\int \left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^3\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx

Evaluating the integral on the right-hand side, we get:

\int \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} dx = x\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} - 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^3\exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right<br/> **Q&A: Does the Integral Converge?**

A: The given integral is $ \int_0^\infty \exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx$, where $m>0$ . This integral is significant in the field of mathematics and physics, as it can be used to model real-world phenomena, such as the behavior of physical systems over an infinite time period.

A: As $x$ approaches infinity, the term $\frac{m}{\sqrt{2}x}$ approaches zero. This is because the denominator $x$ becomes very large, causing the fraction to approach zero. Therefore, as $x$ approaches infinity, the integrand approaches:

\exp\left{-\left[\frac{x}{\sqrt{2}}\right]^2\right}

This is a Gaussian function, which decays exponentially as $x$ approaches infinity.

A: To determine whether the integral converges, we need to evaluate the integral over the infinite interval. One way to do this is to use the method of integration by parts.

A: The method of integration by parts is a technique used to integrate functions that are products of two functions. It involves choosing a function $u$ and its derivative $du$ , and then integrating the product of $u$ and $dv$ .

A: We can apply the method of integration by parts to the given integral by choosing $u = \exp\left\{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right\}$ and $dv = dx$ . Then, we have:

du = -2\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx

v = x

Now, we can integrate by parts:

Evaluating the integral on the right-hand side, we get:

Now, we can integrate by parts again:

\int \exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx = x\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} - 2\sqrt{2}\int \left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{2\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} dx

Evaluating the integral on the right-hand side, we get:

\int \exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx = x\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} - 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{2\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} + 2\sqrt{2}\int \left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{3\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} dx

Evaluating the integral on the right-hand side, we get:

\int \exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx = x\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} - 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{2\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{3\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} + C

where $C$ is the constant of integration.

A: The final result of the integration is:

\int \exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} dx = x\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^2\right} - 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{2\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} + 2\sqrt{2}\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]^{3\exp\left{-\left[\frac{x}{\sqrt{2}}+\frac{m}{\sqrt{2}x}\right]}2\right} + C

where $C$ is the constant of integration.

A: The integral