Question

Multivariable Calculus: derivatives under the integral.

I started a problem but right now I'm not sure which way to go with it. I will attach the problem and what I've done towards a solution.

Answer 1

here's the problem: (my attempt to follow)

Answer 2

\[(d/dx)I(\alpha) = \int\limits_{0}^{infinity} (d/d \alpha)((\ln(1+\alpha^2x^2)/(1+x^2))dx\]

Answer 3

\[\frac{ d }{ dx }I(\alpha) = \int\limits_{0}^{infinity}(\frac{ 1 }{ 1+x^2})(\frac{ 2\alpha x^2 }{ 1+\alpha^2x^2 })\]

Answer 4

\[= 2\alpha \int\limits_{0}^{infinity} \frac{ x^2 }{ (1+x^2)(1+\alpha^2x^2)}\]

Answer 5

not sure where to go from there...

Answer 6

try substitution
x^2 = y
just for partial fractions
(know what partial fractions is?)

Answer 7

yeah I've done partial fractions before... Iet me try...

Answer 8

do not actually calculate dx or change the limits

Answer 9

so for the numerators:
A + B = 0
A(alpha^2) + B = 1

Answer 10

B is \[\frac{ 1 }{ 1-\alpha^2 }\]
A is \[\frac{ -1 }{ 1-\alpha^2 }\]

Answer 11

y = A (1+a^2y) + B (1+y)
yes, those are correct

Answer 12

both A,B are correct
now resubstitute y as x^2

Answer 13

so we get \[-2\alpha \int\limits_{0}^{infinity}(\frac{ 1 }{ (1-\alpha^2)(1+x^2) })+2\alpha \int\limits_{0}^{infinity} \frac{ 1 }{ (1-\alpha^2)(1+\alpha^2x^2) }\]

Answer 14

yes, aren't those easy to integrate using standard formulas ?

Answer 15

integrating with respect to x right?

Answer 16

yes

Answer 17

\[\frac{ -2\alpha }{ 1-\alpha^2 }\int\limits_{0}^{infinity} \frac{ 1 }{ 1+x^2 }\] + \[\frac{ 2\alpha }{ 1-\alpha^2 }\int\limits_{0}^{infinity} \frac{ 1 }{ 1+ \alpha^2x^2 }\]

Answer 18

yes
go on

Answer 19

\[\frac{ -2\alpha }{ 1-\alpha^2 }\tan^{-1} x \] is the first part...

Answer 20

correct

Answer 21

not exactly sure what to do with the alpha^2 in the second part. It looks just like another inverse tangent, but...

Answer 22

it is another tan inverse,
just not in a standard form
we need to bring it in standard form

Answer 23

oh wait got it

Answer 24

the second part is \[\frac{ 1 }{ \alpha}\tan^{-1} (\alpha x)\]

Answer 25

now just set alpha to 1 in order to get the answer to the original question right?

Answer 26

wait a minute...that would create a snafu...

Answer 27

yes, but

Answer 28

you need to simplify things first

Answer 29

working on it now...

Answer 30

actually solve the integrals first
by plugging in upper limit and lower limit and subtracting....

Answer 31

oh right right forgot these were definite integrals.

Answer 32

so the limit as inverse tangent goes to infinity is 1, and when it goes to zero is zero, soo...

Answer 33

\[\frac{ -2\alpha }{ 1-\alpha^2 }+ 1\]

Answer 34

tan inv infinity = pi/2 actually

Answer 35

because tan pi/2 = infinity

Answer 36

oh shucks right... let me redo that...

Answer 37

\[\frac{ \alpha \pi }{ 1-\alpha^2 }+ \frac{ \pi }{ 2}\] is the real answer.

Answer 38

but I still can't plug in alpha = 1...

Answer 39

-2a/ (1-a^2) (pi/2) + 2/(1-a^2) * pi/2

how did u not het this ?

Answer 40

get***

Answer 41

wait what.... I got the first part (think I forgot to put in the - sign though). Let me see what I did on the second part...

Answer 42

i guess you just ignored the 2a/(1=a^2) part

Answer 43

1-a^2 ***

Answer 44

oh right! I forgot to plug it back into the 2nd part...

Answer 45

so now we get...

Answer 46

0???

Answer 47

did you get this part first ?
-2a/ (1-a^2) (pi/2) + 2/(1-a^2) * pi/2

Answer 48

wait how did you get a pi/2 in the denominator of the first part?

Answer 49

both pi/2 are in the numerator separately

Answer 50

Here let me spell things out a bit. I'm kinda confused: \[\frac{ -2\alpha }{ 1-\alpha^2 }(\frac{ \pi }{ 2}-\frac{ 1 }{ \alpha }\frac{ \pi }{ 2 }\alpha)\] is the total we got previously... \[\frac{ -2\alpha }{ 1-\alpha^2 }(\frac{ \pi }{ 2}-\frac{ \pi }{ 2 })\] which is just zero?

Answer 51

where did I go wrong?

Answer 52

how did u get pi*a/2 ??

Answer 53

....-1/a (pi/2)

Answer 54

oh wait yeah oops alpha*infinity is still infinity. whoops.

Answer 55

so you factored out -2a/(1-a^2)
then you should get
(pi/2 - pi/2a) (-2a/(1-a^2))

Answer 56

so its... \[\frac{ -\alpha \pi }{ 1-\alpha^2}(1-\frac{ 1 }{ \alpha })\]

Answer 57

idk how you simplify but in the end i get pi/2
did u get the same ?

Answer 58

\[\frac{ -\alpha \pi }{ 1-\alpha^2 }+ \frac{ \pi }{ 1-\alpha^2 }\]

Answer 59

yes going goo
now ther's a common denominator

Answer 60

good***

Answer 61

so \[\frac{ \pi - \alpha \pi }{ 1 - \alpha^2 }\]

Answer 62

\[\frac{ \pi(1 - \alpha) }{ 1-\alpha^2 }\]

Answer 63

\[\frac{ \pi }{ 1+ \alpha}\]

Answer 64

which is just pi/2. got it

Answer 65

\(\huge \color{blue}\checkmark\)

Answer 66

Thanks again! Appreciate the fact that you just spent an hour saving my skin.

Answer 67

always happy to help :)
welcome ^_^