Question

I need to find the value of k such that the integral from -infinity to infinity of f(x)dx = 1. How can I do this with the function f(x) = k |x|/(x^2 +4) ?

Answer 1

I have tried to integrate to separate out k, but it's not working really well.

Answer 2

hello!

Answer 3

find k such that
\[ \large 1=\int_{-\infty}^{+\infty}\frac{k|x|}{x^2+4}\,dx \]
??

Answer 4

That's what I tried to do, but I don't know how work out the last bits, where I have an exponent on either side and I'm trying to get k.

Answer 5

the function inside the integral is symetric with respect to y-axis. did u notice that?

Answer 6

So I get\[1= 1/2 k \int\limits_{?}^{?}1/(x^2 +4)\]

Answer 7

what do you mean?

Answer 8

oh, I see.

Answer 9

well, I don't know actually, since the variable is x?

Answer 10

\[ \large \frac{|-x|}{(-x)^2+4}=\frac{|x|}{x^2+4} \]
right?

Answer 11

yes

Answer 12

so
\[ \large \int_{-\infty}^{+\infty}\frac{k|x|}{x^2+4}\,dx=
2\int_{0}^{+\infty}\frac{kx}{x^2+4}\,dx \]
\[ \large =2\lim_{b\to+\infty}k\int_0^b\frac{x}{x^2+4}\,dx
=2k\lim_{b\to+\infty}\int_0^b\frac{x}{x^2+4}\,dx \]

Answer 13

putting u=x^2+4 then du=2x dx then x dx=du/2

Answer 14

I've gotten\[1= (1/2) k \int\limits_{-\infty}^{\infty} 1/(x^2 +4) dx\] down to \[1= (1/2) k ((1/2) \tan^-1(x/2) or something like that\]

Answer 15

but wait, you still have a x at the top when you took out the 2 from the intergral

Answer 16

\[ \large =2k\lim_{b\to+\infty}\int_0^b\frac{1}{u}\,\frac{du}{2}=
\frac{2k}{2}\lim_{b\to+\infty}\int_0^b\frac{du}{u} \]

Answer 17

What you last typed is what I have, sort of. I don't understand the 0 to b thing, but I do understand the du part.

Answer 18

and 0 to b covers -infinity to infinity?

Answer 19

yes. since the function u r integrating is symetric with respect to y-axis, the area between the curve and the x-axis is the same to both sides of the y-axis

Answer 20

I have a question: how did you get the 2 outside the integral right before the rest of your work? When I did it, I went right to du and ended up with a (1/2) outside the integral. How did you avoid that?

Answer 21

the total area is twice the area either to the left or the right.

Answer 22

wait, since going from 0 to infinity can be the same as going from 0 to -infinity (same infinite length, I can 'split' the integral? As if I will multiply by two to get the final answer?

Answer 23

yes. it should work if the integral converges.

Answer 24

So, now I am at a similar spot. I know the arc tan trig identity work here for du/u, but how can I solve this out for k?

Answer 25

let mi finish. ok?

Answer 26

okay :D

Answer 27

\[ \large =k\lim_{b\to+\infty}\left(\int_0^1\frac{du}{u}+\int_1^b\frac{du}{u}\right) \]
\[ \large =k\left[\lim_{a\to 0+}\int_a^1\frac{du}{u}+
\lim_{b\to+\infty}\int_1^b\frac{du}{u}\right] \]

Answer 28

\[ \large =k\left[\lim_{a\to0+}(\ln1-\ln a)+
\lim_{b\to+\infty}(\ln b-\ln 1)\right]
=k[-(-\infty)+(+\infty)] \]

Answer 29

done.

Answer 30

wait...k equals...infinity? I'm not sure I understand...,

Answer 31

me neither

Answer 32

r u sure about the |x| in the numerator.

Answer 33

ur, no, if it's 1= k[inifinity + infinity], does k = 1/infinity= facepalm?

Answer 34

yeah, function reads f(x) = k|x|/((x^2) +4)

Answer 35

I'm supposed to solve it so that if it was evaluated from -infinity to infinity, it would equal 1, depending on the correct value of k

Answer 36

\[ \large =k[+\infty+(+\infty)]=k(+\infty) \]

Answer 37

\[ \large =+\infty \]

Answer 38

it is impossible. all the work shows the integral diverges.

Answer 39

hmmm

Answer 40

what if the |x| wasn't in the problem? Would that change things?

Answer 41

b/c I think they don't give impossible problems on the first day of school, lol

Answer 42

then we would have
\[ \large 1=2k\int_0^{+\infty}\frac{1}{x^2+4}\,dx=
2k\lim_{b\to+\infty}\int_0^b\frac{dx}{x^2+4} \]
\[ \large =2k\lim_{b\to+\infty}\frac{1}{2}\arctan\frac{x}{2}\biggr|_0^b \]

Answer 43

yeah, that's what I have

Answer 44

but any way you turn it, you are going to end up multiply k by zero, unless i'm going it wrong

Answer 45

\[ \large 2k\frac{1}{2}\lim_{b\to+\infty}\arctan\frac{b}{2}=k\frac{\pi}{2} \]

Answer 46

that is
\[ \large 1=k\frac{\pi}{2} \]
then
\[ \large k=\frac{2}{\pi} \]

Answer 47

How does arctan(b/2) (input infinity) become...oh

Answer 48

no, what?

Answer 49

\[ \large \lim_{x\to+\infty}\arctan x=\frac{\pi}{2} \]

Answer 50

that is true

Answer 51

where x = b/2?

Answer 52

well if \(b\to+\infty\) then \(b/2\to+\infty\) as well

Answer 53

wait, you have that extra 2 by k again. I thought we got rid of that?

Answer 54

we did!

Answer 55

...so how did it reappear to reduce the other 1/2 from the trig identity? Maybe I missed something.

Answer 56

We had k *lim(b-->+infinity) (du/u), and then we know that du/u = trig identity (1/2) arctan(x/2).

Answer 57

u r forgetting this
\[ \large \int_{-\infty}^{+\infty}\frac{1}{x^2+4}\,dx=2\int_0^{+\infty}
\frac{1}{x^2+4}\,dx \]

Answer 58

yeah, that's the part I'm talking about: we take du and that 2 cancels (2k/2)

Answer 59

so we end up with k(integral) 1/(x^2 +4)

Answer 60

how do we mysteriously get another 2?

Answer 61

Sorry, I really want to understand this. I've got two more problems just like it.