How could a wrong Method solve a very difficult Integral? Magic?

[mario99]

$\displaystyle \frac{a}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-st-\frac{a^2}{4t}}}{\sqrt{t^3}} \ dt = e^{-a\sqrt{s}}$

I will use a wrong method and it will help me to get the above result.

$\displaystyle x = \sqrt{t}$

$\displaystyle dx = \frac{1}{2\sqrt{t}} \ dt$


$\displaystyle \frac{a}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-sx^2-\frac{a^2}{4x^2}}}{x^3} \ 2x \ dx = \frac{a}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-sx^2-\frac{a^2}{4x^2}}}{x^2} \ dx$


$= \displaystyle \frac{a}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{0}^{\infty}\frac{e^{-\left(\sqrt{s}x-\frac{a}{2x}\right)^2}}{x^2} \ dx$


$\displaystyle u = \sqrt{s}x-\frac{a}{2x}$

$\displaystyle du = \left(\sqrt{s} + \frac{a}{2x^2}\right) \ dx$


$\displaystyle \frac{a}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{x^2\left(\sqrt{s} + \frac{a}{2x^2}\right)} \ du = \frac{a}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{\left(\sqrt{s}x^2 + \frac{a}{2}\right)} \ du$

Look above, we still have $\displaystyle x^2 \$ inside the integral.

$\displaystyle ux = \sqrt{s}x^2-\frac{a}{2}$

$\displaystyle x^2 = \frac{ux + \frac{a}{2}}{\sqrt{s}}$

Watch the magic now. Let $\displaystyle u = 0 \$ above. (At here the wrong started.)

$\displaystyle x^2 = \frac{a}{2\sqrt{s}}$

Substitute this value back in the integral.


$\displaystyle \frac{a}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{\left(\sqrt{s}\left[\frac{a}{2\sqrt{s}}\right] + \frac{a}{2}\right)} \ du = \frac{a}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{\left(\frac{a}{2} + \frac{a}{2}\right)} \ du$


$\displaystyle = \frac{a}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{e^{-u^2}}{a} \ du = \frac{1}{\sqrt{\pi}}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}e^{-u^2} \ du$


$\displaystyle = \frac{1}{2}e^{-a\sqrt{s}}\int_{-\infty}^{\infty}\frac{2e^{-u^2}}{\sqrt{\pi}} \ du = \frac{1}{2}e^{-a\sqrt{s}}\text{erf}(u)\bigg|_{-\infty}^{\infty} = \frac{1}{2}e^{-a\sqrt{s}}(1 + 1) = e^{-a\sqrt{s}}$


How could this wrong method give the correct result of the integral? Is it magic or there is an explanation to fill up these mysterious gaps? It worked two times and if no one did it before, Mario will take the credit.

 

[Steven G]

Clearly you can't let u=0 because you can't make a u-sub where u is a function of x and u=0. How can you compute du?!!!
Now you clearly stated that is wrong. I am just pointing out where I see that this would cause a problem.
Why is it working? I am still looking at it. I suspect that you were lucky!

 

[mario99]

Mathematics is so precise so that there is no chance that you do wrong calculations and get the correct result by luck multiple times. This method, it is either valid where we think it is invalid, or it is invalid but there is a rigorous explanation for getting the correct result.

 

[topsquark]

Mathematics is so precise so that there is no chance that you do wrong calculations and get the correct result by luck multiple times. This method, it is either valid where we think it is invalid, or it is invalid but there is a rigorous explanation for getting the correct result.

Pffl. You've clearly never spent any time teaching an Introductory level Mathematics class. Stuff like this happens all the time. From my tutoring, I can pretty much guess that the same is true from at least Jr. High level and up. It even happens on the professional level: the Bohr model is almost completely wrong, but it still gives the correct spectral frequencies. (It really should be scrapped, as it teaches too many wrong things that need to be corrected later on.) And Dirac's prediction of antiparticles, which was spectacularly confirmed, was due to incorrect reasoning. (He couldn't have known that, and he did live long enough to learn why he was wrong and what the correct argument was, but it was still wrong.)

If you are getting a correct result by setting u = 0 then there is some symmetry in the integration that is causing this, not that you have discovered anything that is useful, except perhaps for a small class of integrals. It was luck that you found a shortcut, but as the method does something so clearly incorrect, I'm not going to bother looking at it.

-Dan

 

[mario99]

not that you have discovered anything that is useful, except perhaps for a small class of integrals. It was luck that you found a shortcut, but as the method does something so clearly incorrect, I'm not going to bother looking at it.

I have noticed that this incorrect method always works with integrals of the following form.

$\displaystyle K_{\nu}(z) = \dfrac{1}{2} \left ( \dfrac{z}{2} \right )^{\nu} \int_0^{\infty} \dfrac{e^{-( t + z^2/(4t) )}}{t^{\nu+1}} \, dt$

This means that I have discovered a new method (shorcut) by coincidence to replace the complicated modified Bessel function for solving this specific type of integral. Where should I go exactly to receive the million dollars prize? It is a new invention.

 

[Steven G]

I have noticed that this incorrect method always works with integrals of the following form.

$\displaystyle K_{\nu}(z) = \dfrac{1}{2} \left ( \dfrac{z}{2} \right )^{\nu} \int_0^{\infty} \dfrac{e^{-( t + z^2/(4t) )}}{t^{\nu+1}} \, dt$

This means that I have discovered a new method (shorcut) by coincidence to replace the complicated modified Bessel function for solving this specific type of integral. Where should I go exactly to receive the million dollars prize? It is a new invention.

Let's see a valid proof and then we will all say that you're correct.
Mathematical discovery usually does not come with a monetary prize.

 

[khansaheb]

I have noticed that this incorrect method always works with integrals of the following form.

$\displaystyle K_{\nu}(z) = \dfrac{1}{2} \left ( \dfrac{z}{2} \right )^{\nu} \int_0^{\infty} \dfrac{e^{-( t + z^2/(4t) )}}{t^{\nu+1}} \, dt$

This means that I have discovered a new method (shorcut) by coincidence to replace the complicated modified Bessel function for solving this specific type of integral. Where should I go exactly to receive the million dollars prize? It is a new invention.

Send your invention to the mathematics department of Cambridge or Oxford university for proper recognition.

 

[topsquark]

I have noticed that this incorrect method always works with integrals of the following form.

$\displaystyle K_{\nu}(z) = \dfrac{1}{2} \left ( \dfrac{z}{2} \right )^{\nu} \int_0^{\infty} \dfrac{e^{-( t + z^2/(4t) )}}{t^{\nu+1}} \, dt$

This means that I have discovered a new method (shorcut) by coincidence to replace the complicated modified Bessel function for solving this specific type of integral. Where should I go exactly to receive the million dollars prize? It is a new invention.

I doubt you've found anything new. Your "discovery" is probably simply the symmetry needed to finish off my derivation in your other post. It's beyond me because I don't know Bessel functions all that well, much less the modified ones, but that doesn't mean it isn't a well known principle. As I keep telling the anti-Einsteins on PHF, you need to learn a lot more about what we already know before you go claiming to have found something new. You have barely scratched the surface of Mathematics: I suspect you have no idea about just how much you don't yet know.

-Dan

 

[Steven G]

You have barely scratched the surface of Mathematics: I suspect you have no idea about just how much you don't yet know.

-Dan

The world's best mathematicians have barely scratched the surface of Mathematics.
That actually is amazing. The average high school student knows about the same percentage of all mathematics as the brightest living mathematician.
This goes to show just how much math is out there.
I was at a math conference and a passerby said to their mate 'what can mathematicians possible have a conference for'!