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#### CaptainBlack

##### Well-known member

- Jan 26, 2012

- 890

so i am stuck at one improper integral question. By the way its from further pure 3 module a-level.

first the question asks to substitute y=1/x to transform the the integral lnx^2 / x^3 dx into

integral 2y ln(y) dy. Then it asks to evaluate the integral 2y ln(y) dy with limit (1 , 0) by showing the limiting process.

i did it up to there and i got -0.5 but at last it then asks to find the value of the integral lnx^2 / x^3 dx with limit (infinity , 1). Somehow the answer in the markscheme is 0.5.

can someone please explain how the two integral are connected and how the sign changed?

thanks.

When you make the change of variable you have (for both \(a,b>0\) ):

\[ \int_a^b \frac{\ln(x^2)}{x^3} dx = \int_{1/a}^{1/b} 2y \ln(y) dy \]

So for \(A>0\) :

\[ \int_A^1 \frac{\ln(x^2)}{x^3} dx = \int_{1/A}^{1} 2y \ln(y) dy \]

and so the limits of both sides as \(A\) goes to infinity are equal and hence:

\[ \int_{\infty}^1 \frac{\ln(x^2)}{x^3} dx = \int_{0}^{1} 2y \ln(y) dy =- \int_1^0 2y \ln(y) dy =-(-0.5)=0.5\]

CB

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