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I have to calculate this integral . $$\int _{\frac{1}{4}}^2\:\frac{\ln\left(2x\right)}{x\ln\left(4x\right)}\,dx$$

I have no idea how to start , help someone ? Thanks.

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    $\begingroup$ The integral diverges. $\endgroup$ – Ron Gordon Jan 26 '16 at 18:37
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Note that $\ln(ax) = \ln a + \ln x$. Try $y= \ln x$. Then \begin{align*} \int_{\frac{1}{4}}^2 \frac{\ln 2x}{x\ln 4x} dx &= \int_{-\ln 4}^{\ln 2}\frac{\ln 2+y}{\ln 4 + y} dy. \end{align*} The integral diverge.

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HINT:

$$\int_{\frac{1}{4}}^{2}\frac{\ln(2x)}{x\ln(4x)}\space\text{d}x=$$


Integrate by parts, $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where,

$$f=\ln(2x),\text{d}g=\frac{1}{x\ln(4x)}\space\text{d}x=\text{d}f=\frac{1}{x}\space\text{d}x,g=\ln(\ln(4x))$$


$$\left[\ln(2x)\ln(\ln(4x))\right]_{\frac{1}{4}}^{2}-\int_{\frac{1}{4}}^{2}\frac{\ln(\ln(4x))}{x}\space\text{d}x=$$


Subtitute $u=\ln(4x)$ and $\text{d}u=\frac{1}{x}\space\text{d}x$.

This gives a new lower bound $u=\ln\left(4\cdot\frac{1}{4}\right)=\ln(1)=0$ and upper bound $u=\ln\left(4\cdot2\right)=\ln(8)$:


$$\left[\ln(2x)\ln(\ln(4x))\right]_{\frac{1}{4}}^{2}-\int_{0}^{\ln(8)}\ln(u)\space\text{d}u$$

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Let $x=e^t$, then \begin{align} \int_{\frac{1}{4}}^2\frac{\ln(2x)}{x\ln(4x)}\,dx&=\int_{\ln\left(\frac{1}{4}\right)}^{\ln 2}\frac{\ln 2+t}{e^t(2\ln 2+t)}e^t\,dt\\ &=\int_{\ln\left(\frac{1}{4}\right)}^{\ln 2}\left(1-\frac{\ln 2}{2\ln 2+t}\right)\,dt \end{align} Last integral diverges.

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The integral can be worked out as follows:

$$\int _{\frac{1}{4}}^2\:\frac{\ln\left(2x\right)}{x\ln\left(4x\right)}dx$$ $$\int _{\frac{1}{4}}^2\:\frac{\ln\left(2\right)+\ln\left(x\right)}{\ln\left(4\right)+\ln\left(x\right)}\cdot \frac{1}{x}dx$$

Assume that $\ln x= z$

Then we have that $\frac{1}{x}dx =dz$

Also when $x=\frac{1}{4}$ , $z=-\ln 4$ and $x=2$ , $z=\ln 2$

Hence the integral transforms to $$\int _{-\ln 4}^{\ln 2}\:\frac{\ln\left(2\right)+z}{2\ln\left(2\right)+z} dz$$ $$=\int _{-\ln 4}^{\ln 2}\:\left[1-\frac{\ln\left(2\right)}{2\ln\left(2\right)+z}\right] dz$$ $$=\ln 4+\ln 2 - \ln 2 \ln \left|2\ln\left(2\right)+z\right| _{-\ln 4}^{\ln 2}$$

The last integral as you can see diverges.

Hope this helps.

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$$\int _{ \frac { 1 }{ 4 } }^{ 2 } \: \frac { ln\left( 2x \right) }{ xln\left( 4x \right) } dx=\int _{ \frac { 1 }{ 4 } }^{ 2 }{ \frac { \ln { 2 } +\ln { \left( x \right) } }{ \ln { 4 } +\ln { \left( x \right) } } d\left( \ln { \left( x \right) } \right) } =\int _{ \frac { 1 }{ 4 } }^{ 2 }{ \frac { \ln { 2 } +\ln { \left( x \right) } }{ 2\ln { 2 } +\ln { \left( x \right) } } d\left( \ln { \left( x \right) } \right) } \\ \ln { \left( x \right) =t } \\ \int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \frac { \ln { 2 } +t }{ 2\ln { 2 } +t } dt } =\int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \frac { 2\ln { 2 } +t-\ln { 2 } }{ 2\ln { 2 } +t } dt } =\int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \left( 1-\frac { \ln { 2 } }{ 2\ln { 2 } +t } \right) dt } =\\ =\int _{ \ln { 1/4 } }^{ \ln { 2 } }{ dt } -\ln { 2 } \int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \frac { d\left( 2\ln { 2+t } \right) }{ 2\ln { 2 } +t } = } $$ $$=\ln { 2-\ln { \frac { 1 }{ 4 } } - } \ln { 2 } \left[ \ln { \left| 2\ln { 2+\ln { 2 } } \right| } -\ln { \left| 2\ln { 2+\ln { \frac { 1 }{ 4 } } } \right| } \right] \\ $$

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Notice, $$\int_{1/4}^4\frac{\ln(2x)}{x\ln(4x)}\ dx= \int_{1/4}^4\frac{\ln(4x)-\ln(2)}{x\ln(4x)}\ dx $$ $$=\int_{1/4}^4\left(\frac 1x-\frac{\ln(2)}{x\ln(4x)}\right)\ dx $$ $$=\int_{1/4}^4\frac 1x \ dx-\ln(2)\int_{1/4}^4\frac{d(\ln(4x))}{\ln(4x)}$$ $$=\left[\ln(x)\right]_{1/4}^4-\ln(2)\left[\ln(\ln(4x))\right]_{1/4}^4$$

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