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The Proof is Trivial!

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Original post by FireGarden
If cool is what we want, here's the coolest evaluation of an integral I've seen.

We want to evaluate ∫−∞∞e−ax2 dx=f(a)\displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L][L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so aa must have dimension [L]−2[L]^{-2}, and dxdx of course has dimension [L][L].


Overall the LHS has dimension [1][L][1][L]. Thus f(a)f(a) has these dimensions too, so f(a)=kaf(a)=\frac{k}{\sqrt{a}} where kk is a dimensionless constant. In particular, f(1)=πf(1)=\sqrt{\pi} which defines this constant, hence f(a)=πaf(a)=\sqrt{\frac{\pi}{a}}

Ah, I've had this hinted at by Siklos, but he didn't actually give an example - that is really really neat :biggrin:

EDIT: it's actually nicer than the DUTIS way, I think!
(edited 9 years ago)
Original post by FireGarden
If cool is what we want, here's the coolest evaluation of an integral I've seen.

We want to evaluate ∫−∞∞e−ax2 dx=f(a)\displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L][L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so aa must have dimension [L]−2[L]^{-2}, and dxdx of course has dimension [L][L].


Overall the LHS has dimension [1][L][1][L]. Thus f(a)f(a) has these dimensions too, so f(a)=kaf(a)=\frac{k}{\sqrt{a}} where kk is a dimensionless constant. In particular, f(1)=πf(1)=\sqrt{\pi} which defines this constant, hence f(a)=πaf(a)=\sqrt{\frac{\pi}{a}}


that is just beautiful. I knew you could use dimensional analysis to check forms of equations and build them up, but to actually get f(a) like that is wow!

I think you could use a similar idea for the arctanx integral then.
Original post by FireGarden
If cool is what we want, here's the coolest evaluation of an integral I've seen.

We want to evaluate ∫−∞∞e−ax2 dx=f(a)\displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L][L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so aa must have dimension [L]−2[L]^{-2}, and dxdx of course has dimension [L][L].


Overall the LHS has dimension [1][L][1][L]. Thus f(a)f(a) has these dimensions too, so f(a)=kaf(a)=\frac{k}{\sqrt{a}} where kk is a dimensionless constant. In particular, f(1)=πf(1)=\sqrt{\pi} which defines this constant, hence f(a)=πaf(a)=\sqrt{\frac{\pi}{a}}


That very nice, but can it be made rigorous?
Original post by james22
That very nice, but can it be made rigorous?

You can DUTIS to check the answer, I suppose, but that's no easier than the original :P
Original post by james22
That very nice, but can it be made rigorous?


Well, I'm no expert but it is rigourous. Another way to look at it is to take the 'units' as being some factor to do substitutions with. In the original integral, you can sub x↦kx,a↦ak2x\mapsto kx, a\mapsto\frac{a}{k^2}, then you'll get the functional equation f(a)=1kf(ak2)f(a)=\dfrac{1}{k}f(\dfrac{a}{k^2}) , which I guess* leads to the unique solution f(a)=f(1)af(a)=\dfrac{f(1)}{\sqrt{a}}.

*I guess comes from the initial note - I'm no expert! I don't know much about functional equations at all, but it will surely have some theory that shows this is entirely rigourous.
(edited 9 years ago)
Original post by FireGarden
If cool is what we want, here's the coolest evaluation of an integral I've seen.

We want to evaluate ∫−∞∞e−ax2 dx=f(a)\displaystyle\int_{-\infty}^{\infty} e^{-ax^2}\ dx = f(a).

We will proceed with dimensional analysis. Suppose x represents some length, so it has dimension [L][L]. The exp function is required to have dimensionless input (as it is dimensionless itself), so aa must have dimension [L]−2[L]^{-2}, and dxdx of course has dimension [L][L].


Overall the LHS has dimension [1][L][1][L]. Thus f(a)f(a) has these dimensions too, so f(a)=kaf(a)=\frac{k}{\sqrt{a}} where kk is a dimensionless constant. In particular, f(1)=πf(1)=\sqrt{\pi} which defines this constant, hence f(a)=πaf(a)=\sqrt{\frac{\pi}{a}}


:zomg: I never thought Dimensional Analysis could be used like that.
(edited 9 years ago)
Reply 2846
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Original post by jjpneed1
In case you're interested in how I did it, here's a hint:

Spoiler




Could you post this calculation, please? I can't see it. :no:
Original post by ζ(s)
Could you post this calculation, please? I can't see it. :no:


Spoiler

Reply 2848
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Original post by jjpneed1

Spoiler

I don't think the works because, if I recall, one of the integrals that you end up getting isn't simple at all.
Original post by ζ(s)
I don't think the works because, if I recall, one of the integrals that you end up getting isn't simple at all.


It certainly works. I just checked my working and the only non-simple integral is in fact the integral in the last problem which you have to integrate indefinitely this time. So just give a series solution for it (essentially the last problem, but without plugging in limits) which should make it the integral requiring the least amount of work
Reply 2850
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Original post by jjpneed1
It certainly works. I just checked my working and the only non-simple integral is in fact the integral in the last problem which you have to integrate indefinitely this time. So just give a series solution for it (essentially the last problem, but without plugging in limits) which should make it the integral requiring the least amount of work
Could you post the solution, please, if you don't mind? I'm not sure how you have done it with the series.
Original post by ζ(s)
Could you post the solution, please, if you don't mind? I'm not sure how you have done it with the series.


Spoiler



Apologies for any errors that may arise.. took me quite a while to do
Reply 2852
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Original post by jjpneed1
...


Brilliant stuff! It never dawned on me to ignore the non-closed form and that it would disappear in the end due to the initial condition. Even when you spelled it out as a hint, I was like 'what sorcery is this?' Thanks for typing this up.
Problem 9001:

Prove that:

∫0∞ln⁡x(1+x2)2 dx=−π4\displaystyle \int_0^\infty \frac {\ln x} {(1+x^2)^2}\,\mathrm d x = - \frac {\pi} 4

Edit: Made it a 'show that' question instead.
(edited 9 years ago)
Reply 2854
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Original post by Tarquin Digby
Problem 9001:

Prove that:

∫0∞ln⁡x(1+x2)2 dx=−π4\displaystyle \int_0^\infty \frac {\ln x} {(1+x^2)^2}\,\mathrm d x = - \frac {\pi} 4

Edit: Made it a 'show that' question instead.

Spoiler

(edited 9 years ago)
Original post by ζ(s)

Spoiler




O.O

Alternatively use integration by parts on ∫01(1−x2)ln⁡(x)(1+x2)2dx \displaystyle \int_0^1 \frac{(1 - x^2)\ln(x)}{(1 + x^2)^2} {dx} and the result is immediate. Though your way looks cooler I admit
Reply 2856
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Original post by jjpneed1

O.O

Alternatively use integration by parts on ∫01(1−x2)ln⁡(x)(1+x2)2dx \displaystyle \int_0^1 \frac{(1 - x^2)\ln(x)}{(1 + x^2)^2} {dx} and the result is immediate. Though your way looks cooler I admit


If we want cool, how about defining I(α)=∫0∞ln⁡xα2+x2  dx\displaystyle I(\alpha) = \int_{0}^{\infty}\frac{\ln{x}}{ \alpha^2+x^2}\;{dx} then putting x↦αx \displaystyle x \mapsto \alpha x.
Original post by ζ(s)
If we want cool, how about defining I(α)=∫0∞ln⁡xα2+x2  dx\displaystyle I(\alpha) = \int_{0}^{\infty}\frac{\ln{x}}{ \alpha^2+x^2}\;{dx} then putting x↦αx \displaystyle x \mapsto \alpha x.


Pretty average tbh :wink:

Problem 471
Determine all sets of non-negative integers x,y x, y and zz that satisfy 2x+3y=z2 2^x + 3^y = z^2 .

Problem 472
Evaluate ∫0∞ln⁡(1+x)ln⁡(1+x2)x3dx \displaystyle \int_0^{\infty} \frac{\ln(1 + x)\ln(1 + x^2)}{x^3} dx

Last ones for a while now while I sit some tasty exams, gl on that integral, good job if someone gets it out before exams finish :smile:
(edited 9 years ago)
Why are the problems up to the 9000s..?
Original post by FireGarden
In particular, f(1)=πf(1)=\sqrt{\pi} which defines this constant


Where is this coming from? I don't see how you know this unless you know how to evaluate the integral with a=1a=1 in the usual fashion.

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