Question

Running into trouble while differentiating an improper integral...

Answer 1

Live preview's not working... can't see how my latex is showing up..

Answer 2

The method of dierentiation under the integral sign, due originally to Leibniz, concerns integrals
depending on a parameter, such as R 1
0
x
2
e
tx dx. Here t is the extra parameter. (Since x is the
variable of integration, x is not a parameter.) In general, we might write such an integral as
(1.1) Z b
a
f(x; t) dx;

Answer 3

Ok it's back.. So, let \[f(t)\]be a continuous function such that \[\int_{-\infty}^{\infty} t*|f(t)|dt\]exists and is finite.

Answer 4

Let\[g(x)=\int_{x}^{\infty}(t-x)*f(t)*dt\]find g'(x) and g"(x). So for g'(x) i have..

Answer 5

\[g'(x) = \frac{d}{dx}(\int_{x}^{\infty}(t-x)*f(t)*dt)=\frac{d}{dx}(\int_{x}^{0}(t-x)f(t)dt+\int_{0}^{\infty}(t-x)f(t)dt)\].. Sorry slow tex typer..

Answer 6

Then I did the following:
\[g'(x)=\frac{d}{dx}(\int_{x}^{0}t*f(t)dt)-\frac{d}{dx}(\int_{x}^{0}x*f(t)dt)\]

Answer 7

i see no error yet.

Answer 8

Sorry page crashed on me.. and finally..

Answer 9

Using the fundamental theorem of calc on these, I got:
\[0-x*f(x)*1 - \frac{d}{dx}(x*\int_{x}^{0}f(t)*dt)=-x*f(x)-(1*\int_{x}^{0}f(t)dt \space + \space x*f(x))\]

Answer 10

Which finally is:
\[-2x*f(x)-\int_{x}^{0}f(t)dt\]
So is this the answer?

Answer 11

I wasn't really expecting to see an integral in the end, so could I be doing something wrong?

Answer 12

no

Answer 13

there is an integral in the end

Answer 14

Ok, thanks, can you just double check the steps?

Answer 15

start with ...

\[g(x)=\int_{x}^{\infty}(t-x)*f(t)*dt\]

\[=\int_{x}^{\infty}t*f(t)*dt-\int_{x}^{\infty}x*f(t)*dt\]
\[=\int_{x}^{\infty}t*f(t)*dt-x\int_{x}^{\infty}f(t)*dt\]

now differentiate

Answer 16

That's actually how i did it, was just trying to skip some steps because typing in latex is a bit of a pain. So what I got after that step was:

\[\frac{d}{dx}(\int_{x}^{0} t*f(t)*dt+\int_{0}^{\infty} t*f(t)*dt)-\frac{d}{dx}(\int_{x}^{0}x*f(t)*dt+\int_{0}^{\infty}x*f(t)*dt)\]

Answer 17

After that I end up with:
\[\frac{d}{dx}(\int_{x}^{0}t*f(t)*dt \space )-\frac{d}{dx}(\int_{x}^{0}x*f(t)*dt \space )\]

Answer 18

no

Answer 19

what about the \[\int\limits_{0}^{\infty}x\cdot f(t)dt\]

Answer 20

Finally, sorry i crashed.. Oh you're right..

Answer 21

\[\frac{d}{dx}\left(\int\limits_{x}^{0}tf(t)dt \right )-\frac{d}{dx}\left(\int\limits_{x}^{0}xf(t)dt +\int\limits_{0}^{\infty}xf(t)dt\right )\]

Answer 22

yea, sorry I forgot about the product rule in that one.

Answer 23

yep...

\[\frac{d}{dx}\left(\int\limits_{x}^{0}tf(t)dt \right )-\frac{d}{dx}\left(x\int\limits_{x}^{0}f(t)dt +x\int\limits_{0}^{\infty}f(t)dt\right )\]

Answer 24

you made a small mistake

Answer 25

yes

Answer 26

\[-x*f(x)-(x*(-f(x))+1*\int_{x}^{0}f(t)*dt)-(x*0+\int_{0}^{\infty}f(t)*dt)\]

Answer 27

Gonna delete the other response, otherwise will get confused when writing this down lol, thanks.

Answer 28

so what is your final answer?

Answer 29

\[=0-\int_{x}^{0}f(t)*dt-\int_0^\infty f(t)*dt=-\int_x^\infty f(t)*dt\]

Answer 30

Is that correct?

Answer 31

that's it

Answer 32

Wow.. thanks Zarkon. Never thought I'd have to do math again... getting rusty.

Answer 33

no problem

Answer 34

Will have a bunch more questions in a few more minutes probably, meanwhile gonna find the 2nd derivative of this thing, that one should be much easier.

Answer 35

yep...should be ;)