MHB 8.8.16 LCC 206 Integral at infinity

[karush]

$\large{8.8.16} $
$\tiny\text{LCC 206 Integral at infinity}$
$$I=\int_{0}^{\infty}\frac{x}{\sqrt[5] {x^2 +1}} \,dx= \infty \\$$
$\text{presume just taking the limit
makes the } \\
x\implies\infty \\
\text{thus the integral goes to } \infty$

$\tiny\text{ Surf the Nations math study group}$

 

[alyafey22]

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.

 

[karush]

$$\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} &
\left(x-1\right)du&={} \ d{x}& x&={1-e^{u}}
\end{align} \\
\text{I proceeded but?!? }$$

 

[alyafey22]

$$\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} &
\left(x-1\right)du&={} \ d{x}& x&={1-e^{u}}
\end{align} \\
\text{I proceeded but?!? }$$

Sorry, I meant use subsritution to solve your integral.

 

[karush]

$$I=\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} & \left(x-1\right)
du&={} \ d{x}&
x&={1-e^{u}} \
\end{align} \\
\text{however if } x=1 \\
\text{then u is undefined }$$
$$I=\int_{a}^{b} \frac{u}{1-e^{u}}\,du$$

 

[Prove It]

$$I=\int^1_0 \frac{\ln(1-x)}{x} = - \frac{\pi^2}{6}$$
$$\begin{align}
\displaystyle
u& = {\ln\left({1-x}\right)} & \left(x-1\right)
du&={} \ d{x}&
x&={1-e^{u}} \
\end{align} \\
\text{however if } x=1 \\
\text{then u is undefined }$$
$$I=\int_{a}^{b} \frac{u}{1-e^{u}}\,du$$

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?

 

[alyafey22]

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 $u = x^2+1$

 

[Greg]

Using the Taylor series for $\log(1-x)$,

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

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

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