# Solving an Integral involving a probability density function

• A

TL;DR Summary
A certain function ##f(x)## represents a probability density function. This means that the integral over the definition range of ##x## must be at least equal to a constant. However, I cannot solve the integral.
In an article written by Richard Rollleigh, published in 2010 entitled The Double Slit Experiment and Quantum Mechanics, he argues as follows:

"For something to be predictable, it must be a consistent measurement result. The positions at which individual particles land on the screen are not consistent: each particle could land in any bright fringe. Positions are not predictable. What is consistent is the probability of each particle’s landing at any position, i. e. the probability density function (pdf) of each particle’s position. The pdf of position is just the double slit interference pattern illustrated in Figure 3. It is reproduced any time you repeat the experiment and it is predicted by Equation 2."

$$\displaystyle I={\frac {I_{{\max}}{\lambda}^{2}}{{\pi }^{2}{a}^{2} \left( \sin \left( \theta \right) \right) ^{2}} \left( \cos \left( {\frac {\pi \,d\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2} \left( \sin \left( {\frac {\pi \,a\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2}}$$
If equation 2 represents a pdf, then the following integral:
$$\displaystyle \int_{-\infty }^{\infty }\!{\frac {{\lambda}^{2}}{{\pi }^{2}{a}^{2} \left( \sin \left( \theta \right) \right) ^{2}} \left( \cos \left( {\frac {\pi \,d\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2} \left( \sin \left( {\frac {\pi \,a\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2}}\,{\rm d}\theta$$
must be equal to a constant. I just can't solve this integral. Who can help me?

If equation 2 represents a pdf, then the following integral:
$$\displaystyle \int_{-\infty }^{\infty }\!{\frac {{\lambda}^{2}}{{\pi }^{2}{a}^{2} \left( \sin \left( \theta \right) \right) ^{2}} \left( \cos \left( {\frac {\pi \,d\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2} \left( \sin \left( {\frac {\pi \,a\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2}}\,{\rm d}\theta$$
must be equal to a constant. I just can't solve this integral. Who can help me?
Try Wolfram Alpha?

PS The range of ##\theta## is ##[-\dfrac{\pi}{2}, +\dfrac{\pi}{2}]##. The substitution ##u = \dfrac{\pi\sin \theta}{\lambda}## is probably a good start. Then try Wolfram Alpha.

• DaveE and Dale
PeroK@ Can you tell me why the range of ##\theta## is ##\displaystyle [- \frac{\pi}{2}, \, + \frac{\pi}{2}]##?

Integration by substitution doesn't work because the derivative of ##u## is again a function of ##\theta##.

PeroK@ Can you tell me why the range of ##\theta## is ##\displaystyle [-\pi /2,\,+\pi /2]##?
Because ##\theta## is the diffraction angle, with ##\theta = 0## representing the line directly between the slits.
Integration by substitution doesn't work because the derivative of ##u## is again a function of ##\theta##.
It looks tricky. My guess is that the integral is independent of the width of the slits (##a##), but that ##I_{max}## depends on the slit separation (##d##) and the wavelength (##\lambda##).

The next try would be a numerical computation.

• Klystron
My guess is that ... that ##I_{max}## depends on the slit separation (##d##) and the wavelength (##\lambda##).
It must.

For large separation you get two single slit patterns - and the formula may even break down.

For large ##\lambda## you get almost no diffraction and two points of high intensity.

\begin{align*}
\displaystyle \int_{-\pi/2 }^{\pi/2 }&\;{\frac {{\lambda}^{2}}{{\pi }^{2}{a}^{2} \left( \sin \left( \theta \right) \right) ^{2}} \left( \cos \left( {\frac {\pi \,b\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2} \left( \sin \left( {\frac {\pi \,a\sin \left( \theta \right) }{\lambda}} \right) \right) ^{2}}\,{\rm d}\theta\\[10pt]
&\stackrel{u=\sin \theta\, , \,\omega =\pi/\lambda }{=}\dfrac{b\omega }{a\omega }\int_{-1}^{1}
\dfrac{\cos^2(b\omega u)}{b\omega u}\dfrac{\sin^2(a\omega u)}{a\omega u} \dfrac{du}{\sqrt{1-u^2}}\\[10pt] &\stackrel{v=a\omega u}{=} 2(a\omega )^2\int_{0}^{a\omega }\cos^2\left(\dfrac{b}{a}v\right)\dfrac{\sin^2(v)}{v^2\sqrt{(a\omega)^2-v^2}}\,dv\\[10pt]
&= 2(a\omega )^2 \int_0^{a\omega } \underbrace{\dfrac{\cos^2\left(\dfrac{b}{a}v\right)\sin(v^2)}{v^2}}_{\text{continuous}} \underbrace{\dfrac{1}{\sqrt{(a\omega)^2-v^2}}}_{\text{integrable and }\geq 0}\,dv\\[10pt]
&=2(a\omega )^2 \,\dfrac{\cos^2\left(\dfrac{b}{a}\xi\right)\sin(\xi^2)}{\xi^2} \,\int_0^{a\omega }\dfrac{dv}{\sqrt{(a\omega)^2-v^2}}<\infty \text{ for some }\;\xi\in [0,a\omega ]
\end{align*}

Last edited:
There should be another factor of ##u^2## on the denominator there. I set ##k = \frac{\pi a}{\lambda}## and ##u = k\sin \theta##, so that ##d\theta = \frac{du}{\sqrt{k^2 - u^2}}##. We need to show that the following integral converges:
$$\int_0^k \frac{\cos^2\big (\frac d a u \big )}{\sqrt{k^2 - u^2}} \frac{\sin^2 u}{u^2} \ du$$Now, the following function is bounded (by ##1##):
$$\cos^2\big (\frac d a u \big )\frac{\sin^2 u}{u^2}$$And:
$$\int_0^k \frac{1}{\sqrt{k^2 - u^2}} = \frac \pi 2$$And that's all we need.

The simpler version of the integral can be done using Wolfram Alpha, using:
$$\int_0^1 \frac{\cos^2(kx)}{\sqrt{1-x^2}} dx = \frac \pi 4 [J_0(2k) + 1]$$Where ##J_0## is a Bessel function of the first kind.

The more complicated version times out on the free version (perhaps I should invest in one of these maths engines?):
$$\int_0^1 \frac{\cos^2(kx)\sin^2(lx)}{x^2\sqrt{1-x^2}} dx = ?$$

There should be another factor of ##u^2## on the denominator there. I set ##k = \frac{\pi a}{\lambda}## and ##u = k\sin \theta##, so that ##d\theta = \frac{du}{\sqrt{k^2 - u^2}}##. We need to show that the following integral converges:
$$\int_0^k \frac{\cos^2\big (\frac d a u \big )}{\sqrt{k^2 - u^2}} \frac{\sin^2 u}{u^2} \ du$$Now, the following function is bounded (by ##1##):
$$\cos^2\big (\frac d a u \big )\frac{\sin^2 u}{u^2}$$And:
$$\int_0^k \frac{1}{\sqrt{k^2 - u^2}} = \frac \pi 2$$And that's all we need.
Yes, I wrote it in the editor here and missed the ##u## in the trig terms. I have corrected it for the sake of completeness, thanks.

But we have a singularity in both factors, one at each end of integration. How can we get boundedness or continuity?

But we have a singularity in both factors, one at each end of integration. How can we get boundedness or continuity?
The singularity in ##\dfrac{sin \ u}{u}## is removable. And ##\dfrac{1}{\sqrt{1 - u^2}}## is integrable on ##[0, 1]##. That's enough, surely?

• fresh_42
The singularity in ##\dfrac{sin \ u}{u}## is removable. And ##\dfrac{1}{\sqrt{1 - u^2}}## is integrable on ##[0, 1]##. That's enough, surely?
Yes, that is sufficient. (See problem 27 in my challenge thread of last month).

• PeroK
Since I was mostly into complex analysis, I would straightaway substitute the complex identities for sine and cosine and se what happens. Exponentials are much easier to handle than trig functions.

Since I was mostly into complex analysis, I would straightaway substitute the complex identities for sine and cosine and se what happens. Exponentials are much easier to handle than trig functions.
That won't save you. You will end up in exp / polynomial.

Since I was mostly into complex analysis, I would straightaway substitute the complex identities for sine and cosine and se what happens. Exponentials are much easier to handle than trig functions.
Even the simplified integral throws up a Bessel function.

The simpler version of the integral can be done using Wolfram Alpha, using:
$$\int_0^1 \frac{\cos^2(kx)}{\sqrt{1-x^2}} dx = \frac \pi 4 [J_0(2k) + 1]$$Where ##J_0## is a Bessel function of the first kind.

The more complicated version times out on the free version (perhaps I should invest in one of these maths engines?):
$$\int_0^1 \frac{\cos^2(kx)\sin^2(lx)}{x^2\sqrt{1-x^2}} dx = ?$$

Using trig identities yields$$\begin{split} \int_0^1 \frac{\cos^2(kx)\sin^2(lx)}{x^2\sqrt{1-x^2}} dx &= \int_0^1 \frac{\sin^2((k+l)x) + \sin^2((k-l)x) + 2\sin((k+l)x)\sin((k-l)x)}{4x^2\sqrt{1-x^2}}\,dx \\ &= \frac{f(k+l,k+l) + f(k-l,k-l) + 2f(k+l,k-l)}{4} \end{split}$$ where $$f(a,b) = \int_0^1 \frac{\sin(ax)\sin(bx)}{x^2\sqrt{1-x^2}}\,dx.$$ WolframAlpha can do the case $a = b$ with a result in terms of Bessel and Struve functions, but cannot do the
$a \neq b$ case. However, it does give $$\int_0^1 \frac{\sin(ax) \sin(bx)}{\sqrt{1-x^2}}\,dx = \frac{\pi}{4} (J_0(a-b) - J_0(a+b))$$ and differentiating $f$ twice with respect to $a$ or $b$ would put the integrand into that form.

• PeroK
The integral can be done analytically to obtain an infintie series of Bessel functions. It's best not to substitute $x =\sin\theta$ but to start from
$$\sin(a\sin\theta) = 2\sum_{k=0}^\infty J_{2k+1}(a) \sin((2k+1)\theta)$$ (Abramowitz & Stegun, 9.1.43). Hence $$\begin{split} \cos(a\sin\theta)\sin(b\sin\theta) &= \frac{\sin((a+b)\sin\theta) - \sin((a-b)\sin\theta)}{2}\\ &= \sum_{k=0}^\infty (J_{2k+1}(a+b) - J_{2k+1}(a-b))\sin((2k+1)\theta) \\ &= \sum_{k=0}^\infty A_k \sin(2k+1)\theta \end{split}$$ which leaves us with $$\begin{split} \int_{-\pi/2}^{\pi/2} \frac{1}{\sin^2\theta} \cos^2(a\sin\theta)\sin^2(b\sin\theta)\,d\theta &= \sum_{k=0}^\infty\sum_{l=0}^\infty A_kA_l \int_{-\pi/2}^{\pi/2} \frac{\sin((2k+1)\theta) \sin((2l+1)\theta)}{\sin^2 \theta}\,d\theta \\ &= 2\sum_{k=0}^\infty\sum_{l=0}^\infty A_kA_l \int_{0}^{\pi/2} U_{2k}(\cos \theta) U_{2l}(\cos \theta)\,d\theta \end{split}$$ where $U_n(\cos\theta) = \sin ((n+1)\theta)/\sin\theta$ is a Chebyshev Polynomial of the Second Kind, $$U_n(x) = \sum_{m=0}^{\lfloor n/2 \rfloor} (-1)^m \frac{(n-m)!}{m!(n-2m)!} (2x)^{n-2m}$$ (Abramowitz & Stegun, 22.3.7). Thus the problem reduces to integrating powers of $\cos\theta$.

• PeroK