8.8.16 LCC 206 Integral at infinity

In summary, the conversation discusses the use of substitution and Taylor series to solve integrals. The first integral, $\int^1_0 \frac{\ln(1-x)}{x}\,dx$, is given as an example and is shown to be equal to $-\frac{\pi^2}{6}$. The conversation then moves on to discuss the integral $I=\int_{0}^{\infty}\frac{x}{\sqrt[5] {x^2 +1}} \,dx$, and suggests using the substitution $u = x^2+1$ to solve it. Finally, the use of the Taylor series for $\log(1-x)$ is mentioned to evaluate the integral.
  • #1
karush
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$\large{8.8.16} $
$\tiny\text{LCC 206 Integral at infinity}$
$$I=\int_{0}^{\infty}\frac{x}{\sqrt[5] {x^2 +1}} \,dx= \infty \\$$
$\text{presume just taking the limit
makes the } \\
x\implies\infty \\
\text{thus the integral goes to } \infty$

$\tiny\text{ Surf the Nations math study group}$
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  • #2
You can't just take the limit to prove divergence. Consider the integral

$$\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$

Regardless of $$\lim_{x \to 1} \frac{\ln(1-x)}{x } = -\infty$$

To solve the integral I suggest using a substitution.
 
  • #3
$$\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} &
\left(x-1\right)du&={} \ d{x}& x&={1-e^{u}}
\end{align} \\
\text{I proceeded but?!? }$$
 
  • #4
karush said:
$$\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} &
\left(x-1\right)du&={} \ d{x}& x&={1-e^{u}}
\end{align} \\
\text{I proceeded but?!? }$$

Sorry, I meant use subsritution to solve your integral.
 
  • #5
$$I=\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} & \left(x-1\right)
du&={} \ d{x}&
x&={1-e^{u}} \
\end{align} \\
\text{however if } x=1 \\
\text{then u is undefined }$$
$$I=\int_{a}^{b} \frac{u}{1-e^{u}}\,du$$
 
  • #6
karush said:
$$I=\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} & \left(x-1\right)
du&={} \ d{x}&
x&={1-e^{u}} \
\end{align} \\
\text{however if } x=1 \\
\text{then u is undefined }$$
$$I=\int_{a}^{b} \frac{u}{1-e^{u}}\,du$$

Edit, I made a mistake, fixing now...

$\displaystyle \begin{align*} \int_0^1{ \frac{\ln{\left( 1 - x \right)} }{x} \,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u = 1 - x \implies \mathrm{d}u = -\mathrm{d}x \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(1) = 0 \end{align*}$, giving

$\displaystyle \begin{align*} \int_0^1{ \frac{\ln{ \left( 1 - x \right) }}{x}\,\mathrm{d}x } &= -\int_0^1{ \frac{\ln{ \left( 1 - x \right) }}{x }\,\left( -1 \right) \,\mathrm{d}x } \\ &= -\int_1^0{ \frac{\ln{(u)}}{1 - u}\,\mathrm{d}u } \\ &= \int_0^1{ \frac{\ln{(u)}}{1 - u}\,\mathrm{d}u } \\ &= \int_0^1{ \frac{\ln{(u)}}{u}\,\left( \frac{u}{1 - u} \right) \,\mathrm{d}u } \end{align*}$

Now use integration by parts with $\displaystyle \begin{align*} U = \frac{u}{1 - u} \implies \mathrm{d}U = \frac{1\,\left( 1 - u \right) - u \,\left( -1 \right)}{\left( 1 - u \right) ^2}\,\mathrm{d}u = \frac{1}{\left( 1 - u \right) ^2 }\,\mathrm{d}u \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}V = \frac{\ln{(u)}}{u} \,\mathrm{d}u \implies V = \frac{\left[\ln{(u)} \right] ^2}{2} \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_0^1{ \frac{\ln{(u)}}{u}\,\left( \frac{u}{1 - u} \right) \,\mathrm{d}u } &= \left[ \frac{u\,\left[ \ln{(u)} \right] ^2 }{2\,\left( 1 - u \right) } \right]_0^1 - \int_0^1{ \frac{\left[ \ln{(u)} \right] ^2 }{2\,\left( 1 - u \right) ^2 }\,\mathrm{d}u } \end{align*}$

Can you continue?
 
Last edited:
  • #7
I gave the integral

$$\int^1_0 \frac{\ln(1-x)}{x}\,dx$$

as an example. It is NOT an elementary integral.

To solve the integral

$$I=\int_{0}^{\infty}\frac{x}{\sqrt[5] {x^2 +1}} \,dx$$

Use the substitution $u = x^2+1$
 
  • #8
Using the Taylor series for $\log(1-x)$,

\(\displaystyle \int_0^1\dfrac{\log(1-x)}{x}\,dx=-\int_0^1\left(\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{n}\right)\,dx\)

\(\displaystyle =\left.\left(-\sum_{n=1}^{\infty}\dfrac{x^n}{n^2}\right)\right|_0^1=-\zeta(2)+0=-\dfrac{\pi^2}{6}\)

where $\zeta(s)$ is the Riemann zeta function.
 

FAQ: 8.8.16 LCC 206 Integral at infinity

1. What is "8.8.16 LCC 206 Integral at infinity"?

"8.8.16 LCC 206 Integral at infinity" refers to a specific mathematical concept known as an integral at infinity, which is often encountered in calculus and other higher-level mathematics.

2. What does "LCC 206" refer to in this context?

"LCC 206" is likely a reference to a specific course or class, such as Linear Control and Computation 206, in which the concept of integral at infinity may be taught or discussed.

3. How is integral at infinity different from a regular integral?

An integral at infinity is a type of improper integral, meaning that one or both of the limits of integration are infinite. This is different from a regular integral, which has finite limits of integration.

4. What is the significance of the date "8.8.16" in this context?

The date "8.8.16" likely refers to the date on which the concept of integral at infinity was discussed or introduced in the course "LCC 206." It may also be a date associated with a specific problem or example involving an integral at infinity.

5. Why is understanding integral at infinity important for scientists?

Integral at infinity is a fundamental concept in mathematics, and it has many applications in various scientific fields such as physics, engineering, and economics. It allows scientists to solve problems involving infinite quantities, making it a crucial tool in many areas of research and analysis.

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