Drawing conclusions by looking at integral

[WWCY]

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!

Homework Equations

The Attempt at a Solution

 

[David Dyer]

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

 

[vela]

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 $I=0$?

 

[WWCY]

Thank you both for replying!

To Dr Dave:

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:

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?

 

[Ray Vickson]

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!

Homework Equations

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]

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.

 

[Ray Vickson]

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!

Homework Equations

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.

 

[WWCY]

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!