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Verification of $(-frac{1}{2})!$

Proving $int_{0}^{infty} mathrm{e}^{-x^2} dx = frac{sqrt pi}{2}$A difficult integral $int_0^infty mathrm{d}tfrac{1}{t}frac{1}{t-s-mathrm{i}epsilon}frac{1}{X}lnfrac{1-X}{1+X} $Substitution of Differentials without IntegralsHow to find $int frac{e^{-x^2}}{x^2 + 1} dx$?Evaluate $lim_{n to infty} int_0^1 frac{dx}{x^n + 1}$Strange Limit in Proof of the Fresnel IntegralHow to transform $- int_{0}^{infty} ln(1 - x e^{-x}) mathrm{d}x$ into $sum_{n=1}^{infty} frac{n!}{n^{n+2}}$?$ int_2^{infty}frac{1}{xsqrt{x^2-4}}dx$ convergence verificationdivergence of $sum_{n=3}^infty frac{sqrt{n}+2}{n-2}$ verification/ alternative methodA naive change of integration limit

0

$begingroup$

I was working on a proof for $(-frac{1}{2})!$ and my issue was with converting my bounds from variable to variable. As you will see, I kept my bounds in terms of $t$ throughout the calculation, but switched the bounds as one of my final steps.

$$int_{t=0}^{t=infty} t^{n}e^{-t}dt$$

$$int_{t=0}^{t=infty} t^{-frac{1}{2}}e^{-t}dt$$
$u=t^{-frac{1}{2}}$ $rightarrow$ $t=u^{-2},dt=-2u^{-3}du$
$$int_{t=0}^{t=infty} ue^{-u^{-2}}cdot -2u^{-2}u^{-1} du$$
$u^{-2}=v^{2}rightarrow v=u^{-1},u=v^{-1}rightarrow dv=-u^{-2}$
$$2int_{t=0}^{t=infty} e^{-v^{2}} dv$$
Recall: $$int_{0}^{infty} e^{-x^{2}} dx = frac{sqrt{pi}}{2}$$
Now for the bounds. I have kept them in terms of $t$, and now, after all of the substitutions, I can relate $v$ to $t$. I have $u=t^{-frac{1}{2}}$, and I also have $u=v^{-1}$, which gives me $v^{-1}=t^{-frac{1}{2}}$ which means that $v=t^{frac{1}{2}}$. When $t=0$, I get: $v=0^{frac{1}{2}}=0$, and when $t=infty$, I get: $v=infty^{frac{1}{2}}=infty$, so my bounds stay the same.
Finally: $$2int_{0}^{infty} e^{-v^{2}} dv=sqrt{pi}$$

integration proof-verification improper-integrals factorial

share|cite|improve this question

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What's the problem? Seems fine.
  $endgroup$
  – Hobbyist
  Mar 19 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is my method for determining the bounds in terms of $t$ to $v$ ok?
  $endgroup$
  – ItIsLastThursday
  Mar 19 at 14:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's pretty much the standard proof that $Gamma(tfrac12)=sqrtpi$ that you normally see. There's no problem with your limits.
  $endgroup$
  – Hobbyist
  Mar 19 at 14:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you try to achieve with the first substitution? Don't you think $t=v^2$ is enough?
  $endgroup$
  – Diger
  Mar 19 at 14:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realized that afterwards, but whatever.
  $endgroup$
  – ItIsLastThursday
  Mar 19 at 14:47
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

I was working on a proof for $(-frac{1}{2})!$ and my issue was with converting my bounds from variable to variable. As you will see, I kept my bounds in terms of $t$ throughout the calculation, but switched the bounds as one of my final steps.

$$int_{t=0}^{t=infty} t^{n}e^{-t}dt$$

$$int_{t=0}^{t=infty} t^{-frac{1}{2}}e^{-t}dt$$
$u=t^{-frac{1}{2}}$ $rightarrow$ $t=u^{-2},dt=-2u^{-3}du$
$$int_{t=0}^{t=infty} ue^{-u^{-2}}cdot -2u^{-2}u^{-1} du$$
$u^{-2}=v^{2}rightarrow v=u^{-1},u=v^{-1}rightarrow dv=-u^{-2}$
$$2int_{t=0}^{t=infty} e^{-v^{2}} dv$$
Recall: $$int_{0}^{infty} e^{-x^{2}} dx = frac{sqrt{pi}}{2}$$
Now for the bounds. I have kept them in terms of $t$, and now, after all of the substitutions, I can relate $v$ to $t$. I have $u=t^{-frac{1}{2}}$, and I also have $u=v^{-1}$, which gives me $v^{-1}=t^{-frac{1}{2}}$ which means that $v=t^{frac{1}{2}}$. When $t=0$, I get: $v=0^{frac{1}{2}}=0$, and when $t=infty$, I get: $v=infty^{frac{1}{2}}=infty$, so my bounds stay the same.
Finally: $$2int_{0}^{infty} e^{-v^{2}} dv=sqrt{pi}$$

integration proof-verification improper-integrals factorial

share|cite|improve this question

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What's the problem? Seems fine.
  $endgroup$
  – Hobbyist
  Mar 19 at 14:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is my method for determining the bounds in terms of $t$ to $v$ ok?
  $endgroup$
  – ItIsLastThursday
  Mar 19 at 14:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's pretty much the standard proof that $Gamma(tfrac12)=sqrtpi$ that you normally see. There's no problem with your limits.
  $endgroup$
  – Hobbyist
  Mar 19 at 14:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you try to achieve with the first substitution? Don't you think $t=v^2$ is enough?
  $endgroup$
  – Diger
  Mar 19 at 14:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I realized that afterwards, but whatever.
  $endgroup$
  – ItIsLastThursday
  Mar 19 at 14:47
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I was working on a proof for $(-frac{1}{2})!$ and my issue was with converting my bounds from variable to variable. As you will see, I kept my bounds in terms of $t$ throughout the calculation, but switched the bounds as one of my final steps.

$$int_{t=0}^{t=infty} t^{n}e^{-t}dt$$

$$int_{t=0}^{t=infty} t^{-frac{1}{2}}e^{-t}dt$$
$u=t^{-frac{1}{2}}$ $rightarrow$ $t=u^{-2},dt=-2u^{-3}du$
$$int_{t=0}^{t=infty} ue^{-u^{-2}}cdot -2u^{-2}u^{-1} du$$
$u^{-2}=v^{2}rightarrow v=u^{-1},u=v^{-1}rightarrow dv=-u^{-2}$
$$2int_{t=0}^{t=infty} e^{-v^{2}} dv$$
Recall: $$int_{0}^{infty} e^{-x^{2}} dx = frac{sqrt{pi}}{2}$$
Now for the bounds. I have kept them in terms of $t$, and now, after all of the substitutions, I can relate $v$ to $t$. I have $u=t^{-frac{1}{2}}$, and I also have $u=v^{-1}$, which gives me $v^{-1}=t^{-frac{1}{2}}$ which means that $v=t^{frac{1}{2}}$. When $t=0$, I get: $v=0^{frac{1}{2}}=0$, and when $t=infty$, I get: $v=infty^{frac{1}{2}}=infty$, so my bounds stay the same.
Finally: $$2int_{0}^{infty} e^{-v^{2}} dv=sqrt{pi}$$

integration proof-verification improper-integrals factorial

share|cite|improve this question

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364

$endgroup$

share|cite|improve this question

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364

share|cite|improve this question

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364

asked Mar 19 at 14:40

ItIsLastThursdayItIsLastThursday

364