[quote]
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.[/quote]
When you make the change of variable you have:
\[ \int_a^b \frac{\ln(x^2)}{x^3} dx = \int_{1/a}^{1/b} 2y \ln(y) dy \]
So:
\[ \int_A^1 \frac{\ln(x^2)}{x^3} dx = \int_{1/A}^{1} 2y \ln(y) dy \]
and so the limits of both sides if they 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
