Improper Integral - # 6

[Jason76]

Pretty sure, you need partial fractions for this one.

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v}\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{(v)(v - 1)}\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{(v)(v - 1)} = \dfrac{A}{v} + \dfrac{B}{v - 1}\)


\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = A(v - 1) + B(v)\)


\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = Av - A + Bv\)


\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = - A + Av + Bv\)


\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = - A + v(A + B)\)

\(\displaystyle 2= -A\)

\(\displaystyle 0 = A + B\)

Subsitiuting

\(\displaystyle 2= -A\)

\(\displaystyle 0 = (-2) + B\)

so

\(\displaystyle B = 2\) while \(\displaystyle A = -2\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = \int_{2}^{\infty} \dfrac{-2}{v} + \dfrac{2}{v - 1}\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = -2\ln|v| + 2\ln|v - 1|\) evaluated at 2 (lower bound) and infinity (upper bound)

 

[Ishuda]

Use the log rules, i.e. the difference of log functions is the log of the quotient and others, to re-write your expression as
\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = \ln{(\frac{v-1}{v})^2}|_2^{\infty}\) = 2 * \(\displaystyle \ln{(2)} = \ln{(4)}\) ~ 2 * 0.693147

Edit to correct dumb mistake: Highlight text to see rest of corrections :wink:

 

[Jason76]

Use the log rules, i.e. the difference of log functions is the log of the quotient and others, to re-write your expression as
\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = \ln{(\frac{v-1}{v})^2}|_2^{\infty}= \ln{(2)}\) ~ 0.693147

The book says the answer is the \(\displaystyle \ln(4)\)

 

[stapel]

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = -2\ln|v| + 2\ln|v - 1|\) evaluated at 2 (lower bound) and infinity (upper bound)

Use the log rules, i.e. the difference of log functions is the log of the quotient and others, to re-write your expression as \(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = \ln{(\frac{v-1}{v})^2}|_2^{\infty}\)

The book says the answer is the \(\displaystyle \ln(4)\)

I think the responder accidentally dropped the "2":

. . . . .\(\displaystyle \displaystyle{ \int\, \frac{2}{v^{2}\, -\, v}\, dv\, =\, -2\ln\left|v\right|\, +\, 2\ln\left|v\, -\, 1\right|\, =\, 2\left(\ln\left|v\, -\, 1\right|\, -\, \ln\left|v\right|\right)\, =\, 2 \ln\left|\frac{v\, -\, 1}{v}\right|\, =\, \ln\left(\left(\frac{v\, -\, 1}{v}\right)^2\right) }\)

By retaining that multiplier and then moving it inside the log, what happens when you evaluate at the limits?

 

[HallsofIvy]

A couple of notes on "writing style"

Pretty sure, you need partial fractions for this one.

Yes! Good call!

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v}\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{(v)(v - 1)}\)

You have dropped the "dv" here

\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{(v)(v - 1)} = \dfrac{A}{v} + \dfrac{B}{v - 1}\)

You have dropped the "dv" on the left and dropped the whole integral on the right!
What you mean is that \(\displaystyle \frac{2}{v(v-1)}= \frac{A}{v}+ \frac{B}{v- 1}\) inside the integral.

\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = A(v - 1) + B(v)\)

Now it has become just silly! Surely you saw that? You have taken the denominator out of the integral!
You mean, of course, that, multiplying both sides of \(\displaystyle \frac{2}{v(v- 1)}= \frac{A}{v}+ \frac{B}{v- 1}\) by v(v- 1) you get
\(\displaystyle 2= A(v- 1)+ Bv\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = Av - A + Bv\)


\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = - A + Av + Bv\)


\(\displaystyle \int_{2}^{\infty} \dfrac{2 }{1} = - A + v(A + B)\)

\(\displaystyle 2= -A\)

\(\displaystyle 0 = A + B\)

These last two lines are correct! But before "2= -A" and "0= A+ B" You should not have had the "\(\displaystyle \int_2^\infty\)"
Another way to do this is to note that 2= A(v- 1)+ Bv must be true for all v.
In particular, if v= 0, 2= A(0- 1)+ B(0) or 2= -A. If v= 1, 2= A(0)+ B(1) or B= 2.

Subsitituting

\(\displaystyle 2= -A\)

\(\displaystyle 0 = (-2) + B\)

so

\(\displaystyle B = 2\) while \(\displaystyle A = -2\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = \int_{2}^{\infty} \dfrac{-2}{v} + \dfrac{2}{v - 1}\)

Now this is a correct statement!

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = -2\ln|v| + 2\ln|v - 1|\) evaluated at 2 (lower bound) and infinity (upper bound)

You are going to have difficulty adding or subtracting "infinity" but it is relatively easy to "multiply or divide" (a real number times infinity is infinity, a real number divided by infinity is 0).

So first write -2 ln|v|+ 2 ln|v- 1| as 2(ln(v- 1)- ln(v))= 2(ln|(v- 1)/v| When v= 2 that will be 2 ln(1/2) but more importantly as v goes to infinity, (v- 1)/v goes to 1 and ln(1)= 0. (Notice, "as v goes to infinity" NOT "when v= infinity"!)

 

[Ishuda]

The book says the answer is the \(\displaystyle \ln(4)\)

Didn't you see those invisible 2's. I guess I will have to make it clearer
2 ln(2) = ln(4) ~ 2 * 0.693147

Just go back and highlight the answer.

 

[lookagain]

Subsitiuting [sic]

\(\displaystyle 2= -A\)

\(\displaystyle 0 = (-2) + B\)

so

\(\displaystyle B = 2\) while \(\displaystyle A = -2\)

\(\displaystyle \int_{2}^{\infty} \dfrac{2 dv}{v^{2} - v} = \int_{2}^{\infty} \dfrac{-2}{v} + \dfrac{2}{v - 1}\)

Now this is a correct statement!

It's almost correct. The "dv" was left off of the right-hand side of the equation.


\(\displaystyle \displaystyle\int_{2}^{\infty} \bigg(\dfrac{-2}{v} + \dfrac{2}{v - 1}\bigg)dv\)