Convergence with P-test

[rayfodder]

Hi,

I need help understanding with the portion I highlited in pink. How was the function on the right created in both cases and why is it greater/less than the other one.

 

[Deleted member 4993]

Hi,

I need help understanding with the portion I highlited in pink. How was the function on the right created in both cases and why is it greater/less than the other one.

Cannot see your image.

 

[DrPhil]

Hi,

I need help understanding with the portion I highlited in pink. How was the function on the right created in both cases and why is it greater/less than the other one.

[The highlighting makes it very difficult to read.]

For the first one, I think the 2 in the denominatoor on the right shouldn't be there. The inequality then is true because they have multiplied by a number (x-1) which is between 1 and 2. Since that "2" is a constant, it doesn't affect the convergence of the integral.

For the second case, they are trying to find an integrable function which is everywhere less than the integrand you are testing, but which can be shown to diverge. For u in the range (0,1], the denominator on the right is greater than the denominator on the left, so the right side is smaller.

How do you know what to use for a test function? Experience, diligence, and luck.

 

[stapel]

For other viewers, this is the text from the image:

Problem. Determine whether

. . . . .$\displaystyle \displaystyle{\int_1^3\, \dfrac{dx}{(x\, -\, 1)\sqrt{3\, -\, x}}}$

converges or diverges.

Proof. First, note that this integral is improper because the function has a vertical asymptote at BOTH endpoints so we need to split the integral at some interior point. Then,

. . . . .$\displaystyle \displaystyle{\int_1^3\, \dfrac{dx}{(x\, -\, 1)\sqrt{3\, -\, x}}\, =\, \int_1^2\, \dfrac{dx}{(x\, -\, 1)\sqrt{3\, -\, x}}\, +\, \int_2^3\, \dfrac{dx}{(x\, -\, 1)\sqrt{3\, -\, x}}}$

so we will examine each integral separately.

For the second integral, consider that $\displaystyle \displaystyle{\dfrac{1}{(x\, -\, 1)\sqrt{3\, -\, x}}\, \leq\, \dfrac{1}{2\sqrt{3\, -\, x}}}$ for $\displaystyle 2\, <\, x\, <\, 3$. Thus,

. . . . .$\displaystyle \displaystyle{\int_2^3\, \dfrac{dx}{(x\, -\, 1)\sqrt{3\, -\, x}}}$

converges if

. . . . .$\displaystyle \displaystyle{\int_2^3\, \dfrac{dx}{2\sqrt{3\, -\, x}}}$

converges. Then we use a $\displaystyle u$-substitution with $\displaystyle u\, =\, 3\, -\, x$ ($\displaystyle x\, =\, 3$ implies $\displaystyle u\, =\, 0, x\, =\, 2$ implies $\displaystyle u\, =\, 1$, and $\displaystyle -du\, =\, dx$). So

. . . . .$\displaystyle \displaystyle{\int_2^3\, \dfrac{dx}{2\sqrt{3\, -\, x}}\, =\, \dfrac{1}{2}\, \int_1^0\, \dfrac{-du}{u^{\frac{1}{2}}}\, =\, \dfrac{1}{2}\, \int_0^1\, \dfrac{du}{u^{\frac{1}{2}}}}$

which converges by the power rule for proper integrals. Thus the second interval converges.

Now we consider the first integral. Do a $\displaystyle u$-substitution with $\displaystyle u\, =\, x\, -\, 1$ ($\displaystyle x\, =\, 1$ implies $\displaystyle u\, =\, 0$, $\displaystyle x\, =\, 2$ implies $\displaystyle u\, =\, 1$, and $\displaystyle du\, =\, dx$.) Thus,

. . . . .$\displaystyle \displaystyle{\int_1^2\, \dfrac{dx}{(x\, -\, 1)\sqrt{3\, -\, x}}\, =\, \int_0^1\, \dfrac{dx}{u\, \sqrt{2\, -\, u}}}$

Then, for $\displaystyle 0\, <\, u\, <\, 1$, we have that $\displaystyle \dfrac{1}{u\, \sqrt{2\, -\, u}}\, >\, \dfrac{1}{u\, \sqrt{2}}$. Since

. . . . .$\displaystyle \displaystyle{\int_0^1\, \dfrac{1}{u\, \sqrt{2}}\, =\, \dfrac{1}{\sqrt{2}}\, \int_0^1\, u^{-1}\, du}$

diverges, we know that

. . . . .$\displaystyle \displaystyle{\int_0^1\, \dfrac{dx}{u\, \sqrt{2\, -\, u}}}$

diverges. Thus, the first integral diverges.

Since the first integral diverges and the second integral converges, the entire integral diverges.

 

[rayfodder]

thanks

thanks a bunch guys