# Integrate lnx/10^x?

Mar 12, 2016

mistake

#### Explanation:

$\int \frac{\ln x}{10} ^ x \mathrm{dx}$ can also be written as $\int \left(\ln x\right) \times {10}^{- x} \mathrm{dx}$.

Now, we can use the formula for integral of product

$\int u \cdot v \cdot \mathrm{dx} = u \cdot v - \int \left(v \cdot \mathrm{du}\right)$, where $u = \ln x$

As such, we have $\mathrm{du} = \left(\frac{1}{x}\right) \mathrm{dx}$ and let $\mathrm{dv} = {x}^{- 10} \mathrm{dx}$ or $v = {x}^{- 9} / - 9$

Hence, $\int u \cdot v \cdot \mathrm{dx} = \left(- \frac{1}{9}\right) \ln x . {x}^{- 9} - \int \left({x}^{- 9} / - 9\right) \cdot \frac{\mathrm{dx}}{x}$, or

= $\left(- \frac{1}{9}\right) \ln x . {x}^{- 9} + \left(\frac{1}{9}\right) \int {x}^{- 10} \cdot \mathrm{dx}$

= $\left(- \frac{1}{9}\right) \ln x . {x}^{- 9} + \left(\frac{1}{9}\right) {x}^{- 9} / \left(- 9\right) + c$

= $\left(- \frac{1}{9}\right) \ln x . {x}^{- 9} - \left(\frac{1}{81}\right) {x}^{- 9} + c$

= $- \frac{1}{81} \left({x}^{- 9}\right) \left(9 \ln x + 1\right) + c$

Mar 12, 2016

Appears infinite series integral to me.

#### Explanation:

We can use the formula for integral of product of two function $u \left(x\right) \mathmr{and} v \left(x\right)$

$\int u \cdot \mathrm{dv} = u \cdot v - \int v \cdot \mathrm{du}$

(rule can be simply be derived by integrating the product rule of differentiation)
Given integral $\int \ln \left(x\right) / {10}^{x} \cdot \mathrm{dx}$ can be written as
$\int \ln \left(x\right) \times {10}^{- x} \cdot \mathrm{dx}$

Let $u = \ln \left(x\right) \mathmr{and} \mathrm{dv} = {10}^{- x} \cdot \mathrm{dx}$
from first assumption $\mathrm{du} = \frac{1}{x} \cdot \mathrm{dx}$
from the second equality $v = \int {10}^{-} x \cdot \mathrm{dx} = - \frac{1}{\ln} 10 {10}^{-} x + C$

We get $\int \ln \left(x\right) \times {10}^{- x} \cdot \mathrm{dx} = \ln \left(x\right) \cdot \left(- \frac{1}{\ln} 10 {10}^{-} x + C\right) - \int \left(- \frac{1}{\ln} 10 {10}^{-} x + C\right) \cdot \frac{1}{x} \cdot \mathrm{dx}$
Where $C$ is a constant of integration.
$= \ln \left(x\right) \cdot \left(- \frac{1}{\ln} 10 {10}^{-} x + C\right) + \int \frac{1}{\ln} 10 {10}^{-} x \cdot \frac{1}{x} \cdot \mathrm{dx} - \int C \cdot \frac{1}{x} \cdot \mathrm{dx}$
$= \ln \left(x\right) \cdot \left(- \frac{1}{\ln} 10 {10}^{-} x + C\right) + \int \frac{1}{\ln} 10 {10}^{-} x \cdot \frac{1}{x} \cdot \mathrm{dx} - C \cdot \ln | x | + {C}_{2} ,$simplifying
$= \ln \left(x\right) \cdot \left(- \frac{1}{\ln} 10 {10}^{-} x\right) + \frac{1}{\ln} 10 \int {10}^{-} x \cdot \frac{1}{x} \cdot \mathrm{dx} + {C}_{2}$

It reduces to finding the integral of $\int {x}^{-} 1 \cdot {10}^{-} x \cdot \mathrm{dx}$
Again using the above integral by parts formula
Let $u = {x}^{-} 1$ and $\mathrm{dv} = {10}^{- x} \cdot \mathrm{dx}$
$\mathrm{du} = - {x}^{-} 2 \cdot \mathrm{dx}$ and we already have the value for $v$
$\int {x}^{-} 1 \cdot {10}^{-} x \cdot \mathrm{dx} = {x}^{-} 1 \cdot \left(- \frac{1}{\ln} 10 {10}^{-} x + C\right) - \int \left(- \frac{1}{\ln} 10 {10}^{-} x + C\right) \cdot \left(- {x}^{-} 2 \cdot \mathrm{dx}\right)$

1. Inspection reveals it turns out to be finding $\int {10}^{-} x \cdot {x}^{-} 2 \cdot \mathrm{dx}$ and so on.
2. Function $\ln \left(x\right)$ is defined only for $x > 0$
3. The integral appears to be infinite series integral.
Mar 15, 2016

$\left(\ln y\right) \left(\ln \left({\ln}_{10} y\right)\right) - \ln y = \left(\ln y\right) \left(\ln \left({\ln}_{10} y\right) - 1\right)$

Then put in ${10}^{x}$ for $y$

(ln 10^x )(ln(ln_10 10^x)-ln 10^x

#### Explanation:

Let $y = {10}^{x}$

$\ln y = \ln {10}^{x}$

$\ln y = x \cdot \ln 10$

$x = \ln \frac{y}{\ln} 10 = {\ln}_{10} y = {\log}_{10} e \times {\log}_{e} y$
$\therefore \mathrm{dx} = {\log}_{10} e \times \frac{1}{y} \times \mathrm{dy}$

$\int \frac{\ln \left({\ln}_{10} y\right)}{y} \times {\log}_{10} e \times \frac{1}{y} \times \mathrm{dy}$
=int (ln(ln_10 y))/y^2xxlog_10exxdy ; u=ln(ln_10 y)=ln(1/ln10 *lny), dv=1/y

$\mathrm{du} = \frac{1}{\ln \frac{y}{\ln} 10} \cdot \frac{1}{y \ln 10} = \left(\ln \frac{10}{\ln} y\right) \left(\frac{1}{y \ln 10}\right) = \frac{1}{y \ln y}$

$v = \ln y$

$u v - \int v \mathrm{du} \to \left(\ln \left({\ln}_{10} y\right)\right) \ln y - \int \ln y \cdot \frac{1}{y \ln y}$

$\left(\ln y\right) \left(\ln \left({\ln}_{10} y\right)\right) - \int \frac{1}{y}$

$\left(\ln y\right) \left(\ln \left({\ln}_{10} y\right)\right) - \ln y = \left(\ln y\right) \left({\ln}_{10} y - 1\right)$

Then put in ${10}^{x}$ for $y$

ln 10^x (ln(ln_10 10^x)-ln 10^x

$P R O O F :$
$\frac{d}{\mathrm{dy}} \left(\left(\ln y\right) \left(\ln \left({\ln}_{10} y\right) - 1\right)\right)$

f=lny, g=ln(ln_10 y) -1)

$f ' = \frac{1}{y} , g ' = \left(\frac{1}{\ln} _ 10 y\right) \left(\frac{1}{y \ln 10}\right)$

$f g ' + g f '$---> product rule
$\ln y \cdot \left(\frac{1}{\ln} _ 10 y\right) \left(\frac{1}{y \ln 10}\right) + \left(\ln \left({\ln}_{10} y\right) - 1\right) \cdot \frac{1}{y}$

$\ln y \left(\frac{1}{\ln \frac{y}{\ln} 10}\right) \left(\frac{1}{y \ln 10}\right) + \left(\ln \left({\ln}_{10} y\right) - 1\right) \cdot \frac{1}{y}$

$\ln y \left(\ln \frac{10}{\ln} y\right) \left(\frac{1}{y \ln 10}\right) + \left(\ln \left({\ln}_{10} y\right) - 1\right) \cdot \frac{1}{y}$

$\frac{1}{y} + \frac{\ln \left({\ln}_{10} y\right) - 1}{y}$

$\frac{\left(1 + \ln \left({\ln}_{10} y\right) - 1\right)}{y}$

$\frac{\ln \left({\ln}_{10} y\right)}{y}$

$\ln \frac{x}{10} ^ x$--->${\ln}_{10} y = x$ from above