# Drawing conclusions by looking at integral

• WWCY
In summary, the integral ##I## can be expressed as an integral over the entire real line, and a counterexample can be constructed using a symmetric distribution ##f_x## that has two delta-function peaks at ##\pm b##. This leads to a contradiction with the proposed statement that ##I## must be positive if the area under the graph on the negative ##p##-axis is smaller than that on the positive ##p##-axis.

## Homework Statement

I have the following expression

$$I = \int_{-\infty}^{0} f_p(p) \ \big[ pf_x(a - \frac{p}{m}t) \big] dp + \int_{0}^{\infty} f_p(p) \ \big[ pf_x(a - \frac{p}{m}t) \big] dp$$

where ##f_p## and ##f_x## are normalised distributions. In particular, ##f_x## is symmetric about ##x=a##. This also means that ##pf_x(a - \frac{p}{m}t) ## is anti-symmetric about ##p = 0##.

Is there any way through which I can prove (or disprove) that ##I## is positive if
$$\int_{-\infty}^{0} f_p(p) dp < \int_{0}^{\infty} f_p(p) dp$$
ie the area under the graph on the -ve ##p##-axis is smaller than that on the +ve ##p##-axis?

Some guidance regarding how to start is greatly appreciated. Thanks in advance!

## The Attempt at a Solution

I would try to express the (negative) left side wrt the (positive) side. I.E. something like - (int_0^+inf, the integrand will change but you have properties of f and f_x to help you out. PLEASE, let me know if/how that helps.

(Dr.) Dave

Can't you come up with a counterexample where ##f_p(p)## and ##g_p(p) = pf_x(a-\frac pm t)## don't overlap so that ##I=0##?

To Dr Dave:
David Dyer said:
I would try to express the (negative) left side wrt the (positive) side. I.E. something like - (int_0^+inf, the integrand will change but you have properties of f and f_x to help you out. PLEASE, let me know if/how that helps.
hmm, does this not assume that ##f_p## is symmetric about 0 as well? I'm not sure I'm allowed to make such an assumption in my problem if that is the case.

To vela:
vela said:
Can't you come up with a counterexample where ##f_p(p)## and ##g_p(p) = pf_x(a-\frac pm t)## don't overlap so that ##I=0##?
So if I have ##g_p (p)## that asymptotes at about ##p = \pm 5## ,and I have a ##f_p## that has gaussian peaks at ##p = -1## (area of ~ 0.4) and ##p = 20## (area of ~0.6), I can even have a negative value for ##I## despite ##\int_{-\infty}^{0} f_p(p) dp < \int_{0}^{\infty} f_p(p) dp##

Is this right?

WWCY said:

## Homework Statement

I have the following expression

$$I = \int_{-\infty}^{0} f_p(p) \ \big[ pf_x(a - \frac{p}{m}t) \big] dp + \int_{0}^{\infty} f_p(p) \ \big[ pf_x(a - \frac{p}{m}t) \big] dp$$

where ##f_p## and ##f_x## are normalised distributions. In particular, ##f_x## is symmetric about ##x=a##. This also means that ##pf_x(a - \frac{p}{m}t) ## is anti-symmetric about ##p = 0##.

Is there any way through which I can prove (or disprove) that ##I## is positive if
$$\int_{-\infty}^{0} f_p(p) dp < \int_{0}^{\infty} f_p(p) dp$$
ie the area under the graph on the -ve ##p##-axis is smaller than that on the +ve ##p##-axis?

Some guidance regarding how to start is greatly appreciated. Thanks in advance!

## The Attempt at a Solution

Let ##I = I_1 +I_2,## where ##I_1## is the first integral and ##I_2## is the second one. Putting ##p = -p'## in ##I_2## we have
$$I_2 = \int_{\infty}^0 f_p(-p') (-p') f_x\left(a + \frac{p'}{m}t \right) (-dp')\\ = -\int_0^{\infty} f_p(-p') p' f_x\left(a +\frac{p'}{m}t \right) \, dp'.$$
Now change the integration variable back to ##p## and use symmetry of ##f_x##, to get
$$I = \int_0^{\infty} p [f_p(p)-f_p(-p)] f_x\left(a - \frac{p}{m} t \right) \, dp.$$

WWCY
Ray Vickson said:
Let ##I = I_1 +I_2,## where ##I_1## is the first integral and ##I_2## is the second one. Putting ##p = -p'## in ##I_2## we have
$$I_2 = \int_{\infty}^0 f_p(-p') (-p') f_x\left(a + \frac{p'}{m}t \right) (-dp')\\ = -\int_0^{\infty} f_p(-p') p' f_x\left(a +\frac{p'}{m}t \right) \, dp'.$$
Now change the integration variable back to ##p## and use symmetry of ##f_x##, to get
$$I = \int_0^{\infty} p [f_p(p)-f_p(-p)] f_x\left(a - \frac{p}{m} t \right) \, dp.$$

Thank you so much for helping out!

Edit: Does this apply to ##I_1## instead of ##I_2##? I'm am not really able to see why the integration limits for ##I_2## carry the same sign despite the change in variables.

WWCY said:

## Homework Statement

I have the following expression

$$I = \int_{-\infty}^{0} f_p(p) \ \big[ pf_x(a - \frac{p}{m}t) \big] dp + \int_{0}^{\infty} f_p(p) \ \big[ pf_x(a - \frac{p}{m}t) \big] dp$$

where ##f_p## and ##f_x## are normalised distributions. In particular, ##f_x## is symmetric about ##x=a##. This also means that ##pf_x(a - \frac{p}{m}t) ## is anti-symmetric about ##p = 0##.

Is there any way through which I can prove (or disprove) that ##I## is positive if
$$\int_{-\infty}^{0} f_p(p) dp < \int_{0}^{\infty} f_p(p) dp$$
ie the area under the graph on the -ve ##p##-axis is smaller than that on the +ve ##p##-axis?

Some guidance regarding how to start is greatly appreciated. Thanks in advance!

## The Attempt at a Solution

I think the result is incorrect.

First note that you can write your integral ##I## as
$$I = \int_{-\infty}^{\infty} p f_p(p) f_x\left(a- \frac{pt}{m} \right) \, dp$$
Now consider a special case of
$$f_x(w) = \frac{1}{2} \delta(w+b) + \frac{1}{2} \delta(w-b),$$
where ##\delta(u) ## is the Dirac delta-function and ##b > 0##. The delta-function is even, so ##f_x## is also even in ##w##.

Now we get
$$I = \frac{m^2}{2 t^2} \left[(a+b)f_p\left(\frac{m(a+b)}{t} \right) +(a-b) f_p\left(\frac{m(a-b)}{t} \right) \right] \hspace{5ex}(1)$$
By playing with the function ##f_p## and the value of ##b## it will be possible to have ##I < 0## in some cases, but ##I > 0## in other cases.

Of course, you could object that the Dirac delta is not a true, "legitimate" function, but it IS a limit of a sequence of true functions, and so there are true functions that essentially mimic eq. (1), and so give counterexamples to what you hoped.

Last edited:
WWCY
Ray Vickson said:
I think the result is incorrect.

First note that you can write your integral ##I## as
$$I = \int_{-\infty}^{\infty} p f_p(p) f_x\left(a- \frac{pt}{m} \right) \, dp$$
Now consider a special case of
$$f_x(w) = \frac{1}{2} \delta(w+b) + \frac{1}{2} \delta(w-b),$$
where ##\delta(u) ## is the Dirac delta-function and ##b > 0##. The delta-function is even, so ##f_x## is also even in ##w##.

Now we get
$$I = \frac{m^2}{2 t^2} \left[(a+b)f_p\left(\frac{m(a+b)}{t} \right) +(a-b) f_p\left(\frac{m(a-b)}{t} \right) \right] \hspace{5ex}(1)$$
By playing with the function ##f_p## and the value of ##b## it will be possible to have ##I < 0## in some cases, but ##I > 0## in other cases.

Of course, you could object that the Dirac delta is not a true, "legitimate" function, but it IS a limit of a sequence of true functions, and so there are true functions that essentially mimic eq. (1), and so give counterexamples to what you hoped.

I believe I see it clearly now, many thanks for your assistance!