MHB Priyata G integral question from Yahoo Answers

[CaptainBlack]

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

 

[member38644]

Hi Priyata,

Thank you for sharing your question. It seems like you have made some good progress in solving the improper integral. To understand how the two integrals are connected and how the sign changed, let's break down the steps in solving the integral.

First, you have substituted y=1/x to transform the integral lnx^2 / x^3 dx into integral 2y ln(y) dy. This is a common technique used in solving integrals, known as the substitution method. By substituting y=1/x, you have essentially changed the variable from x to y, making the integral easier to solve.

Next, you have evaluated the integral 2y ln(y) dy with limits (1,0) by showing the limiting process. This means that you have taken the limit of the integral as the upper limit approaches 1 and the lower limit approaches 0. This is usually done to check the convergence of the integral. In this case, you have correctly found the limit to be -0.5.

Now, in order to find the value of the integral with limits (infinity, 1), you need to understand the connection between the two integrals. As you can see, the only difference between the two integrals is the limits. In the first integral, the limits are (1,0) and in the second integral, the limits are (infinity,1). However, by using the substitution method, you have essentially made the limits of the two integrals equivalent. This means that the integral with limits (infinity,1) is equal to the integral with limits (1,0) when the limits are taken as infinity and 0 respectively.

To understand the sign change, let's look at the limits of the two integrals. In the first integral, the upper limit is 1 and the lower limit is 0. In the second integral, the upper limit is infinity and the lower limit is 1. When you take the limit of the second integral, the upper limit approaches infinity and the lower limit approaches 0, which is the same as the first integral. However, since the limits are reversed, the sign of the integral changes. This is why the answer in the markscheme is 0.5.

I hope this explanation helps you understand the connection between the two integrals and how the sign changed. Keep up the good work in your studies!