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Evaluation of $\displaystyle \int_{0}^{\frac{1}{2}}\frac{\ln(1-x)}{2x^2-2x+1}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{1}{2}}\frac{\ln(1-x)}{2x^2-2x+1}dx\;,$ Put $1-x=t\;,$ Then $dx=-dt$ and changing limits, We get

$$I = \int_{\frac{1}{2}}^{1}\frac{\ln (t)}{2t^2-2t+1}dt$$

Now Using Trigononmetric substution, Is is very Complex,

Now how can I solve it after that, Help me

Thanks.

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Substitute $t=\frac 1{2y}$

$$I=\int_{\frac 12}^1 \frac{\ln t}{2t^2 - 2t+1} dt =\int_{1}^{\frac 12} \frac{ -\ln (2y)}{\frac{1}{2y^2} -\frac 1y +1} \frac{-1}{2y^2} dy$$

$$= \int_{\frac 12}^1 \frac{-\ln2 -\ln y}{2y^2 -2y+1}dy$$

So $$I= \int_{\frac 12}^1 \frac{-\ln2}{2y^2 -2y+1}dy-I$$

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  • $\begingroup$ Nice and simple, but the last $+I$ should be $-I$, I suppose. $\endgroup$ – mickep Apr 16 '16 at 4:14
  • $\begingroup$ @mickep yes, thank you. I have corrected it :) $\endgroup$ – lEm Apr 16 '16 at 4:14
  • $\begingroup$ Very creative attempt! (+1) $\endgroup$ – user331275 Apr 16 '16 at 4:49
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As $2x^2-2x+1=\dfrac{(2x-1)^2+1}2,$ let $2x-1=\tan y$

$$\int_0^{1/2}\dfrac{\ln(1-x)}{2x^2-2x+1}dx=\int_{-\pi/4}^0\left\{\ln(1-\tan y)-\ln2\right\}dy$$

$$=\int_{-\pi/4}^0\ln(1-\tan y)\ dy-\ln2\int_{-\pi/4}^0\ dy$$

Now using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$ $$I=\int_{-\pi/4}^0\ln(1-\tan y)dy=\int_{-\pi/4}^0\left\{\ln2-\ln(1-\tan y)\right\}dy=\ln2\int_{-\pi/4}^0\ dy-I$$

Can you take it from here?

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Too long for a comment:

Well, to build on your approach, we can consider this integral:

$$I(t)=\int_{1/2}^1 \frac{x^{t-1}}{2t^2-2x+1}$$

Now, we can note an identity involving your integral which is really (due to Mellin Transform)

$$2I'(3)-2I'(2)+I'(0)=\frac{1}{2}$$

This approach may help with your calculation by a lot as you only need to calculate $I(t)$

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