Question

Can somebody guide me through the steps for arriving to the solution of the equation below?

Answer 1

\[\int\limits_{- \infty}^{\infty}\frac{ dx }{ (x ^2+d^2)^{3/2} }=\frac{ 2 }{ d ^{2} }\]Got it from this video: http://www.youtube.com/watch?v=iogzYjhjBCI&list=PL18026EBAD7AB24D6

Answer 2

@ivancsc1996 is 'd' a constant?

Answer 3

I can calculate the antiderivative. The part I get lost is:\[\frac{ x }{ \sqrt{x ^{2}+d ^{2}} }^{\infty}_{-\infty}\](The infinities flying around mean evaluated from -infinity to infinity)

d is a constant

Answer 4

\[\int\limits_{-\infty}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }=2\int\limits_{0}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }\]

Answer 5

Wow that just changed everything. Let me see!

Answer 6

have to break it in two

Answer 7

\[=2\lim_{x \rightarrow \infty}\frac{ x }{ d ^{2}\ \sqrt{x ^{2}+d ^{2}}}=2\left( \frac{ 1 }{ d ^{2} } \right)\]

Answer 8

@pgpilot326 Did you use the formula for integrals given in that form?

Answer 9

Solved it!

Answer 10

yes, i used a table for the integral

Answer 11

Slving the limit was the hardest till I saw I had to use the power rule before l'hopitals

Answer 12

Thanks!

Answer 13

@ivancsc1996 I still have a question, is the 'd' a constant or not?

Answer 14

Yes

Answer 15

ok

Answer 16

isn't symmetry beautiful?

Answer 17

lol you watched the vid didnt u @pgpilot326

Answer 18

or are u referring to the symmetry of the function?

Answer 19

symmetry of functions, but in general too

Answer 20

of that function, i should say

Answer 21

About that. I know that this \[\int\limits\limits_{-\infty}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }=2\int\limits\limits_{0}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }\]equation is correct because it makes sense with the symetry of the problem. But I only gave you the function, how can you know that \[\int\limits\limits_{0}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }=\int\limits\limits_{- \infty}^{0}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }\]from looking at the function? @pgpilot326

Answer 22

Oh I see because of the x^2 the function can never be negative and is symetric on the y axis

Answer 23

let \[f \left( x \right)=\frac{ 1 }{ \left( x ^{2}+d ^{2} \right) }\]then\[f \left( x \right)=f \left( -x \right)\]

Answer 24

exactly

Answer 25

Once again, I have a problem with this.\[\int\limits\limits\limits_{0}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }=\int\limits\limits\limits_{- \infty}^{0}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }\]I can see why they are equal grafically but when I actually solve the integral and use the fundamental theorem and the limits as x aproaches inf and -inf, I get 1=-1, Can you guide me through it? @pgpilot326

Answer 26

you have to stubtract at -infinity

Answer 27

subtract

Answer 28

I mean the antiderivative turns out to be: \[\frac{ x }{ (x^2+d^2)^{1/2} }=F(x)\]I solved that \[F(\infty)=F(-\infty)=1\]And using the fundamental theorem I get that the solution is:\[F(\infty)-F(0)+F(0)-F(-\infty)=1-0+0-1=0\]

Answer 29

Thats the solution for \[\int\limits\limits\limits\limits_{- \infty}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }\]

Answer 30

when we equate that to \[\int\limits\limits\limits\limits_{0}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }+\int\limits\limits\limits\limits_{- \infty}^{0}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }\]

Answer 31

\[\int\limits_{-\infty}^{0}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }=0-\lim_{x \rightarrow -\infty}\frac{ x }{ d ^{2}\sqrt{x ^{2}+d ^{2}} }\]

Answer 32

I am sorry, so basically:\[\int\limits\limits\limits\limits_{0}^{\infty}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }+\int\limits\limits\limits\limits_{- \infty}^{0}\frac{ dx }{ \left( x ^{2}+d ^{2} \right)^{\frac{ 3 }{ 2 }} }=F(\infty)-F(0)+F(0)-F(-\infty)=1-0+0-1=0\]

Answer 33

Wow I am sorry I am divaging please keep on.

Answer 34

from -inf to 0 = 1

Answer 35

Why from -inf to 0=1? wouldn't it be equal to -1?

Answer 36

x on top is negative but sqrt of x squared on bottom is positive

Answer 37

that gives -1 but you're subtracting it from 0. look above at post with integral and limit

Answer 38

Oh yeah, you are right.

Answer 39

http://www.wolframalpha.com/input/?i=lim+as+x-%3E-infinity+of+x%2F%28%28x%5E2%2Bd%5E2%29%5E%281%2F2%29%29 Why does in the second step of the step-by-step solution when they simplify the radicals, they put a minus sign in front of the square root?

Answer 40

@primeralph @pgpilot326 I am sorry, this is my last question to understand the topic.

Answer 41

What is the question exactly?

Answer 42

In the link I put. Just before the comment I tagged you in.