FP1: Improper Integrals

Original post by Lunu

I get how to do the process but I just can't seem to understand why some have finite answers and some are undefined. Can anyone help?

E.g. why does the integral of 1 to infinity of x^-2 have an answer but the integral of 1 to infinity of x^-1/2 not?

It is a matter of convergence - they all have answers, (per se) its just that as the upper limit tends to infinity some integrals converge, others do not.
It also depends on what restrictions you place on the domain as to whether the integral converges.

(edited 11 years ago)

Reply 3

OP

I'm still really confused. Sorry I'm awful at FP1. :/

Reply 4

OP

The answer to the first one is -1/n +1
The answer to the second is 2sqrtx -2

The first one has an answer of 1 and the second one doesn't have an answer. Why is that?

(edited 11 years ago)

Reply 5

OP

I get on the first one that dividing by infinity make the answer really small so tends towards zero but I don't get the second one

Reply 6

joostan

Original post by Lunu

I get on the first one that dividing by infinity make the answer really small so tends towards zero but I don't get the second one

Consider the square root of a massive number. Say: 10000000000000000000000000000000000000.
What can you tell me about its square root.
If I were to make this number bigger, what would happen to its square root - then what if it were infinite?

(edited 11 years ago)

Reply 7

Original post by Lunu

I'm still really confused. Sorry I'm awful at FP1. :/

\displaystyle\int_{1}^{\infty} \dfrac{1}{x^{2}} dx = \left[-\dfrac{1}{x}\right]_{1}^{\infty}

Now obviously as x becomes very large,

\frac{1}{x}

becomes very small. You could think about it as

\displaystyle\lim_{a \to \infty} \left[-\dfrac{1}{x}\right]_{1}^{a}= \displaystyle\lim_{a \to \infty} \left(-\dfrac{1}{a} - - \dfrac{1}{1}\right)

and that

\displaystyle\lim_{a \to \infty} \dfrac{1}{a} \to 0

Thus it effectively becomes

\left(0 - - \dfrac{1}{1}\right) = 1

Now think about the other integral

\displaystyle\int_{1}^{\infty} \dfrac{1}{\sqrt{x}} dx = \left[2\sqrt{x} \right]_{1}^{\infty}

Again, think about it as

\displaystyle\lim_{a \to \infty} \left[2\sqrt{x} \right]_{1}^{a}

= \displaystyle\lim_{a \to \infty} \left(2 \sqrt{a} - 2 \sqrt{1} \right)

But

\displaystyle\lim_{a \to \infty} 2 \sqrt{a}

doesn't converge, hence the integral can't be evaluated

(edited 11 years ago)

Reply 8

OP

Original post by joostan

Consider the square root of a massive number. Say: 10000000000000000000000000000000000000.
What can you tell me about its square root.
If I were to make this number bigger, what would happen to its square root - then what if it were infinite?

The square root is very small? It converges to a fixed number?

Reply 9

james22

Original post by Lunu

The square root is very small? It converges to a fixed number?

No it doesn't. Why would it? If the square root function was bounded you can just go 1 above the bound and square that numer to get a contradiction.

Reply 10

OP

Original post by james22

No it doesn't. Why would it? If the square root function was bounded you can just go 1 above the bound and square that numer to get a contradiction.

what?

Reply 11

Original post by Lunu

what?

The point is as x becomes very large

\sqrt{x}

doesn't converge to a limit.

Reply 12

OP

so what happens to it?

Reply 13

davros

Original post by Lunu

I get how to do the process but I just can't seem to understand why some have finite answers and some are undefined. Can anyone help?

E.g. why does the integral of 1 to infinity of x^-2 have an answer but the integral of 1 to infinity of x^-1/2 not?

As far as further maths is concerned, you just have to evaluate the integral as a function of x and then see what happens as x gets larger and larger.

You have the same situation with infinite sums - for example adding up all the numbers of the form 1/n (1/1, 1/2, 1/3 etc) gives you a number that just gets bigger and bigger and doesn't tend to a limit, but if you add up the reciprocals of the square numbers (1/1, 1/4, 1/9 etc) then you do get a finite value.

An integral is just like a great big wibbly-wobbly sum!

Reply 14

Original post by davros

An integral is just like a great big wibbly-wobbly sum!

Wibbly-wobbly timey-wimey sum? :colone:

Reply 15

OP

Original post by davros

As far as further maths is concerned, you just have to evaluate the integral as a function of x and then see what happens as x gets larger and larger.

You have the same situation with infinite sums - for example adding up all the numbers of the form 1/n (1/1, 1/2, 1/3 etc) gives you a number that just gets bigger and bigger and doesn't tend to a limit, but if you add up the reciprocals of the square numbers (1/1, 1/4, 1/9 etc) then you do get a finite value.

An integral is just like a great big wibbly-wobbly sum!

So it's similar to the series stuff in core 2?

Reply 16

james22

Original post by Lunu

what?

Basically, as x gets very large, sqrt(x) also get very large so it will not converge.

Reply 17

Original post by Lunu

so what happens to it?

It continues to get infinitely large - think about the graph

y = \sqrt{x}

it just increases and tends to infinity as x approaches infinity

Reply 18

davros