Question

Multivariable Calculus question:

Hi everyone!
I'm having trouble with the concept of integration with respect to a parameter. Hope someone can help.


This isn't a problem from the MIT course.

Answer 1

Here is the problem in question:

Answer 2

thats DUIS
differentiation under integral sign
heard of it ?

Answer 3

No, I haven't. My professor did an example in class (he probably called it by a different name), but now I'm confused.

Answer 4

can you find the derivative of
1/(a^2+x^2)^2
with respect to 'a'
?

Answer 5

(-4a)/(a^2+x^2)^3 right?

Answer 6

yeah

Answer 7

sorry, i meant to ask you
can you find the derivative of
1/(a^2+x^2)
with respect to 'a'
?
:P

Answer 8

(-2a)/(a^2+x^2)^2 I think...

Answer 9

i'll write the rule by that time,
\(\large F(a)=\int \limits_a^bf(x,a)dx \\ \dfrac{d}{dx}F(a) =\int \limits_a^b \dfrac{\partial}{\partial a}f(x,a)dx\)

Answer 10

and yes! thats correct :)

Answer 11

so we have
\(\large F(a)=\int \limits_a^bf(x,a)dx = \dfrac{\pi}{2a}\\ \dfrac{d}{dx}F(a) =\int \limits_a^b \dfrac{\partial}{\partial a}f(x,a)dx = \dfrac{d}{dx}( \dfrac{\pi}{2a}) = \int_0^\infty \dfrac{2adx}{(x^2+a^2)^2}\)
is this confusing ?

Answer 12

i missed a negative sign :P

Answer 13

oh ouch...

Answer 14

\(\large F(a)=\int \limits_a^bf(x,a)dx = \dfrac{\pi}{2a}\\ \dfrac{d}{dx}F(a) =\int \limits_a^b \dfrac{\partial}{\partial a}f(x,a)dx = \dfrac{d}{dx}( \dfrac{\pi}{2a}) = \int_0^\infty \dfrac{-2adx}{(x^2+a^2)^2}\)

whichpart did u not get ?

Answer 15

I'm just a bit confused...if we're manipulating everything wrt a, then why is everything in dx?

Answer 16

d/dx(pi/2a) would just be zero...

Answer 17

we are JUST INTEGRTATING w.r.t x

and yeah, i made a typo there again! sorry :P

Answer 18

\(\large F(a)=\int \limits_a^bf(x,a)dx = \dfrac{\pi}{2a}\\ \dfrac{d}{da}F(a) =\int \limits_a^b \dfrac{\partial}{\partial a}f(x,a)dx = \dfrac{d}{da}( \dfrac{\pi}{2a}) = \int_0^\infty \dfrac{-2adx}{(x^2+a^2)^2}\)

Answer 19

so the answer would just be (pi/2)ln(a)?

Answer 20

\(\large F(a)=\int \limits_a^bf(x,a)dx \\ \large \dfrac{\pi}{2a}= \int_0^\infty \dfrac{dx}{(x^2+a^2)}\)
just to show the co-relation,

and pi/2 ln a is the left side!
do we need the right side integral ?

Answer 21

oh shucks yeah...

Answer 22

\(\large \int_0^\infty \dfrac{-2adx}{(x^2+a^2)^2} = \dfrac{\pi \ln a }{2}\)

since we are integrating w.r.t x we can take out -2a
\(\large -2a\int_0^\infty \dfrac{dx}{(x^2+a^2)^2} \dfrac{\pi \ln a }{2}\)
i think now you can find the required integral

Answer 23

oh duh...since we're integrating wrt x we can treat a like a constant.

Answer 24

just rearrange a bit and you have the required answer. haha.

Answer 25

yes!

Answer 26

Thank you so much!!!

Answer 27

you're welcome ^_^