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Does it have to be a man? Or can it also be a woman?Her name is Elizabeth Bathory, and if you don’t know who she is or you’ve never heard anything about her, then allow me to regale with a tale about the most prolific female serial killers ever in history.Erzsebet (Elizabeth) Bathory entered the world in Eastern Europe on August 7th, 1560. Daughter of Baron George Bathory and Baroness Anna Bathory. George and Anna were both Bathorys by birth where inbreeding where purity in the noble line was a aristocratic priority.The Bathorys were a powerful Protestant family in Hungary, and included war...

How would you prove this, [math]\displaystyle\int_0^\infty \dfrac{x^{4}e^{x}}{\left(e^{x}-1\right)^{2}} \, \mathrm dx =\dfrac{4\pi ^{4}}{15}?[/math] Let me tell you how I solved it in my undergrad quantum physics course.Wait, but where does quantum physics come in?[math]I = \displaystyle\int\limits_0^{\infty} \frac{x^4 e^x}{(e^x-1)^2} dx[/math]First, let us remove the square in the denominator by integration by parts. This is possible because[math]d \left[ \displaystyle\frac{1}{e^x-1} \right] = -\displaystyle\frac{e^x}{(e^x-1)^2} dx[/math]So we have[math]I = \left[ -\displaystyle\frac{x^4}{e^x-1} \right]_0^{\infty} + 4 \displaystyle\int\limits_0^{\infty} \frac{x^3}{e^x-1} dx[/math]Since the first term vanishes at both limits, we get[math]I = 4 \displaystyle\int\limits_0^{\infty} \frac{x^3}{e^x-1} dx[/math]Now, it turns out that this is an important integral in black body physics.[1]Planck’s law[2] gives us the spectral radiance at frequency [math]\nu[/math] a black body at temperature [math]T[/math]:[math]B_\nu(\nu,T) = \displaystyle\frac{2h\nu^3}{c^2} \frac{1}{e^{\frac{h\nu}{k_BT}}-1}[/math]where[math]h[/math] is the Planck’s constant[3][math]c[/math] is the speed of light[math]k_B[/math] is Boltzmann constant[4]Now, we would like to find the total radiation from the black body, int...

How do I integrate sin(ax) / (e^(2πx)-1) dx, x from 0 to +∞? Let [math]a[/math] be an arbitrary real number.We observe in passing that, as usual, the first order of business with these integrals is the investigation/proof of their con/di/vergence.As such, the given integrand can be decomposed into a product of two functions [math]f(x)\cdot g(x)[/math] such that [math]f(x)[/math] is uniformly bounded on the given interval and is Riemann-integrable on any finite interval while [math]g(x)[/math] vanishes monotonically as [math]x\to+\infty[/math]:As an exercise, we will let the readers unmask [math]f(x)[/math] and [math]g(x)[/math]. But in any case, the given integral, by the Dirichlet’s criterion, converges for any [math]a[/math] (for [math]a=0[/math] the result is trivial).Then:[math]\dfrac{1}{e^{2\pi x} - 1} = e^{-2\pi x}\cdot\dfrac{1}{1-e^{-2\pi x}} \tag{1}[/math]Decompose the rightmost function i...

Note that the [math]1/x[/math] term is essentially the Laplace transform of [math]1[/math], (if [math]x[/math] is our frequency variable) so we can substitute[math]\dfrac{1}{x} = \displaystyle \int_0^{\infty} 1 \cdot e^{-xt} dt \tag*{}[/math]That turns our integral into an iterated integral (making use of Fubini’s theorem):[math]I = \displaystyle \int_0^{\infty} \biggl( \dfrac{1}{1+x^2}- \cos x \biggr) \dfrac{dx}{x} = \int_0^{\infty} \int_0^{\infty} \biggl( \dfrac{1}{1+x^2}- \cos x \biggr) e^{-xt} \ dx \ dt \tag*{}[/math]Let’s switch the order of integration and distribute the exponential:[math]\displaystyle I = \int_0^{\infty} \Biggl( \int_0^{\infty} \dfrac{e^{-xt}}{1+x^2} \ dx - \underbrace{\int_0^{\infty} \cos x e^{-xt} \ dx}_{I_1} \Biggr) \ dt \tag*{}[/math]Note that [math]I_1[/math] is the Laplace transform of [math]\cos x[/math], so it is equal to[math]\mathcal{L}(\cos x) = \dfrac{t}{t^2 + 1} \tag*{}[/math]So, our original integral becomes[math]\displaystyle I = \int_0^{\infty} \Biggl( \int_0^{\infty} \dfrac{e^{-xt}}{1+x^2} \ dx \Biggr) - \dfrac{t}{t^2 + 1} \ dt \tag*{}[/math]For the inside integral, we can make a substitution of [math]u=xt[/math], which gives [math]du = t \ dx \implies dx = du/t[/math].[math]\displaystyle I = \int_0^{\infty} \Biggl( \int_0^{\infty} \dfrac{e^{-u}}{1+\frac{u^2}{t^2}} \dfrac{du}{t} \Biggr) - \dfrac{t}{t^2 + 1} \ dt \tag*{}[/math]Next, we can multiply the numerator and denominator of the inside integral by [math]t^2[/math]:[math]\displaystyle I = \int_0^{\infty} \Biggl( \int_0^{\infty} \dfrac{te^{-u}}{t^2+u^2} \ du \Biggr) - \dfrac{t}{t^2 + 1} \ dt \tag*{}[/math]I’ll stop for a second to explain where we’re going with this. Our goal is to reduce our in...

What are some tips for first time flyers in India? Flying for the first time could be very exciting !Here are some tips to help one making his/her first flight experience good.airSlate SignNow airport at-the-least 90 mins prior to boarding time as this is your first time. You don't need the stress. Airlines are very strict about their boarding time.The points mentioned below are more suited for domestic flights.At the airport:Procedure that is followed at most of the airports in India:1. Show your ticket copy / sms to the security guard at the terminal entrance along with one photo identity proof in original.2. Once you enter the terminal, go to your ...

To highlight a combination of two problem-solving ideas of:generalization andreductioninstead of just one, as if that’s not enough, consider two sibling integrals at once where [math]x[/math] is a (varying) real number carrying radians and [math]a[/math] and [math]b[/math] are arbitrary real numbers whose magnitudes have been fixed ahead of time (call these symbolic or parameterized constants):[math]\displaystyle \mathbb{X} = \int e^{ax}\cos bx dx \tag{1}[/math]and:[math]\displaystyle \mathbb{Y} = \int e^{ax}\sin bx dx \tag{2}[/math]The introduction of the symbolic constants [math]a[/math] and [math]b[/math] can be considered as a (mild) generalization.Move the exponents in (1) and (2) under the differential:[math]\displaystyle \mathbb{X} = \dfrac{1}{a}\int \cos bx de^{ax} \tag{3}[/math][math]\displaystyle \mathbb{Y} = \dfrac{1}{a}\int \sin bx de^{ax} \tag{4}[/math]Integrate by parts (as has been suggested) but only once:[math]\displaystyle \mathbb{X} = \dfrac{e^{ax}\cos bx}{a} + \dfrac{b}{a}\int e^{ax}\sin bx dx \tag{5}[/math][math]\displaystyle \mathbb{Y} = \dfrac{e^{ax}\sin bx}{a} - \dfrac{b}{a}\int e^{ax}\cos bx dx \tag{6}[/math]where we have omitted the cons...

How do I compute [math]\displaystyle\int_{0}^{\infty}\left(\dfrac{\sin x}{x}\right)^2\,dx[/math]? This is probably an outdated technique so the younger generation should take it with a grain of salt but I will show it for completeness.In some, oh well, older Analysis courses (taken by yours truly, :0) there developed, via a series of theorems, a relationship between improper integrals and Riemann-like sums. Using the corresponding theorems, the computation for this particular integral is next to trivial - it takes literally about two lines of effort but setting up the context requires some doing which I will do briefly.If we break the [math][0, +\infty)[/math] interval into an infinite number of line segments o...

For a non-negative real [math]k[/math], what is [math]\sum\limits_{n=0}^{\infty}\dfrac{n^k}{n!}[/math]? I will highlight one way to develop a solution.Do you know any series that may be helpful?Start with:[math]\displaystyle e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \tag{1}[/math]That is the case [math]k = 0[/math] when and if (1) is evaluated at [math]x = 1[/math].Think now - what kind of uniform process can you apply to (1) to start building up the powers of [math]n[/math] in the numerator?Recall that the operation of differentiation with respect to [math]x[/math] of a function of the shape [math]y = x^n[/math] reduces [math]n[/math] by one and “brings [math]n[/math] down”:[math]y_x' = nx^{n-1} \tag*{}[/math]That seems to do exactly what we need but - stop - can we differentiate (1) term by term?No, not really - unless we earn the right to do so by proving that on a certain interval the functions in questions ...

To explain what things mean in the Schrödinger equation, let’s start with something simpler. The Hamiltonian equation for energy for a classical particle is:[math]H=\dfrac{{\bf p}^2}{2m}+V({\bf q}),\tag*{}[/math]where [math]{\bf p}[/math] is the momentum, [math]m[/math] is the particle mass, [math]{\bf q}[/math] is the position, and [math]V({\bf q})[/math] is just a generic potential that is a function of the coordinates. (The same equation with only trivial changes applies to multiparticle systems, too, by the way.)Now let me do a little bit of trivial algebra (no physics, just pure math) and multiply this expression by the unit complex number [math]\psi=e^{i({\bf p}\cdot{\bf q}-Ht)/\hbar}[/math]. As to why I am doing this, it will become evident soon, but for now, just ...