- #1

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limit{n goes to ∞} int_{-∞}^{∞}frac{sqrt(n)}{(x^2+1)^n}=?

Can this calculus be done by complex integral?

I was thinking for a few days and no answer so far : (

Thanks!

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- Thread starter davyyao
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- #1

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limit{n goes to ∞} int_{-∞}^{∞}frac{sqrt(n)}{(x^2+1)^n}=?

Can this calculus be done by complex integral?

I was thinking for a few days and no answer so far : (

Thanks!

- #2

mfb

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With the right integration parts, and a residual at the right point...

As this looks like homework, I moved your thread to the homework section.

- #3

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limit{n goes to ∞} int_{-∞}^{∞}frac{sqrt(n)}{(x^2+1)^n}=?

Can this calculus be done by complex integral?

I was thinking for a few days and no answer so far : (

Thanks!

Hi davy,

You can calculate it by brute force using the Residue Theorem to first compute the integral as a function of [itex]n[/itex], then take the limit. You'll need some Gamma function relationships and know about the rising factorial function and it gets messy. Write it first as:

[tex]\int_{-\infty}^{\infty} \frac{1}{(1+z^2)^n}dz=\int_{-\infty}^{\infty} \left(\frac{1}{(z+i)^n}\right) \left(\frac{1}{(z-i)^n}\right)dz[/tex]

And consider a contour integral over the upper half-disc. And just for now assume the integral over the half-circle arc goes to zero as it's size goes to infinity. Can prove it latter. Then the integral resolves to just computing the residue. However, if you know about the Residue Theorem, then you'll realize you have to compute the general expression for the n-1 derivative of the first expression:

[tex]\frac{d^{(n-1)}}{dz^{(n-1)}}(z+i)^{-n}[/tex]

at [itex]z=i[/itex]. Ok, that's a little tough but you can do it. First try it for say n=5, n=6, the derivatives are easy to compute and resolve to basically the rising factorial function and some other factors. Then use deduction to derive the general expression for any n. Let's see if you can get that far.

Last edited:

- #4

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lim_{n goes to infinity}sqrt(n)*pi*2^(-2n+2)

As I see, I guest it goes to zero as n goes to to infinity.

However, I numerically plot this integrand and it looks delta-like function.

Maybe this integral is finite....I am not sure~~

- #5

Ray Vickson

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limit{n goes to ∞} int_{-∞}^{∞}frac{sqrt(n)}{(x^2+1)^n}=?

Can this calculus be done by complex integral?

I was thinking for a few days and no answer so far : (

Thanks!

If you change variables, first to y = x^2+1, then to u = 1/y, your integral from 0 to ∞ (which is 1/2 of what you want) reduces to a Beta function B(a,b), for some a and b defined in terms of n. The Beta can, in turn be re-expressed in terms of the Gamma function, using the identity

[tex] B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}[/tex]

You end up with a ratio of Gamma functions involving n, multiplied by some computable constants. Now you can take n → ∞ and get a fairly simple limit (using, for example, Stirling's formula to get limits of the Gamma functions for large n).

- #6

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lim_{n goes to infinity}sqrt(n)*pi*2^(-2n+2)

As I see, I guest it goes to zero as n goes to to infinity.

However, I numerically plot this integrand and it looks delta-like function.

Maybe this integral is finite....I am not sure~~

No and no. First it's not hard to just get a ball-park idea of where the limit "might" be going by just numerically integrating say the first 500 terms and plotting them:

Code:

```
mytable = Table[Sqrt[n]*NIntegrate[1/(1 + z^2)^n,
{z, -Infinity, Infinity}], {n, 1, 500}];
mylimit = Graphics[{Red, Line[{{0, Sqrt[Pi]},
{500, Sqrt[Pi]}}]}];
Show[{ListPlot[mytable, Joined -> True], mylimit},
PlotRange -> All]
```

That's Mathematica code and if you don't know it then you have got to go out of your way to find a machine running it and learn it and practice coding it if you want to do math in the 21st century. Anyway, the plot does not seem to be going to zero but that's no guarantee it's not. Remember, just a rough idea.

Secondly, why don't we just then flat-out compute it in Mathematica then figure out how to arrive at that answer. That's not cheating.

Code:

```
In[38]:=
myintegrand = Integrate[1/(1 + z^2)^n,
{z, -Infinity, Infinity}]
Limit[Sqrt[n]*myintegrand[[1]], n -> Infinity]
Out[38]=
ConditionalExpression[(Sqrt[Pi]*Gamma[-(1/2) + n])/
Gamma[n], Re[n] > 1/2]
Out[39]=
Sqrt[Pi]
```

Alright, fine, we got the answer. Now my third point depends if you just want the answer or if you want to learn how to get good in math. If it's the later, then you have got to try really hard to either muscle though my way and arrive at it or Ray's way even if it takes you a week to do so.

- #7

Ray Vickson

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No and no. First it's not hard to just get a ball-park idea of where the limit "might" be going by just numerically integrating say the first 500 terms and plotting them:

Code:`mytable = Table[Sqrt[n]*NIntegrate[1/(1 + z^2)^n, {z, -Infinity, Infinity}], {n, 1, 500}]; mylimit = Graphics[{Red, Line[{{0, Sqrt[Pi]}, {500, Sqrt[Pi]}}]}]; Show[{ListPlot[mytable, Joined -> True], mylimit}, PlotRange -> All]`

That's Mathematica code and if you don't know it then you have got to go out of your way to find a machine running it and learn it and practice coding it if you want to do math in the 21st century. Anyway, the plot does not seem to be going to zero but that's no guarantee it's not. Remember, just a rough idea.

Secondly, why don't we just then flat-out compute it in Mathematica then figure out how to arrive at that answer. That's not cheating.

Code:`In[38]:= myintegrand = Integrate[1/(1 + z^2)^n, {z, -Infinity, Infinity}] Limit[Sqrt[n]*myintegrand[[1]], n -> Infinity] Out[38]= ConditionalExpression[(Sqrt[Pi]*Gamma[-(1/2) + n])/ Gamma[n], Re[n] > 1/2] Out[39]= Sqrt[Pi]`

Alright, fine, we got the answer. Now my third point depends if you just want the answer or if you want to learn how to get good in math. If it's the later, then you have got to try really hard to either muscle though my way and arrive at it or Ray's way even if it takes you a week to do so.

You don't need to learn Mathematica to do math in the 21st century; you can use Maple instead, which is how I got the answer for this problem (while hiding the solution details from the OP).

- #8

mfb

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