0
$\begingroup$

I have the following function which I want to Integrate from zero to infinity:

f[n_] := (-(-1 + Sqrt[1 + y^(2 (n - 2))])^(
    1/3) + (1 + Sqrt[1 + y^(2 (n - 2))])^(1/3))^4/ (y^((n - 2)/3))

for $n=5$ the integral converges

NIntegrate[f[4], {y, 0, Infinity}]
5.31289

from literature we expect that it also should converge for $n>4$, i.e., $n=5,6,7,...$. But Mathematica compute the integral for $n=5,6,7$ with errors and it seems that the integral diverges for $n=6,7$. My question is that should I trust this outcome? Or maybe there is a way that these errors can be fixed and the integrals become finite?

 NIntegrate[f[5], {y, 0, Infinity}, MaxRecursion -> 100]

During evaluation of In[110]:= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In[110]:= NIntegrate::zeroregion: Integration region {{0.,0}} cannot be further subdivided at the specified working precision. NIntegrate assumes zero integral there and on any further indivisible regions.

During evaluation of In[110]:= NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 400 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 442.0378757272312` and 0.0007603109345110766` for the integral and error estimates.

 442.038

NIntegrate[f[6], {y, 0, Infinity}, MaxRecursion -> 100]
3.19445*10^26

NIntegrate[f[7], {y, 0, Infinity}, MaxRecursion -> 100]

During evaluation of In[112]:= NIntegrate::zeroregion: Integration region {{0.,0}} cannot be further subdivided at the specified working precision. NIntegrate assumes zero integral there and on any further indivisible regions.

During evaluation of In[112]:= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

 6.72576*10^51
$\endgroup$
5
  • $\begingroup$ Series[f[5], {y, 0, 1}, Assumptions -> y > 0] implies divergence at y == 0. As does Integrate[f[5], {y, 0, Infinity}], which basically does the same computation. $\endgroup$
    – Michael E2
    Nov 14 '21 at 17:15
  • $\begingroup$ Is the numerator supposed to vanish as $y\rightarrow0$? It doesn't as coded. $\endgroup$
    – Michael E2
    Nov 14 '21 at 17:19
  • $\begingroup$ @MichaelE2 Thank you for your comment, but this is also the case for $f[4]$, but integrating $f[4]$ leads to a finite number. $\endgroup$
    – Kheeyal
    Nov 14 '21 at 17:20
  • $\begingroup$ I get that the integral of f[4] converges, $\sim 16^{1/3} y^{-2/3}$ in which the power is $> -1$. $\endgroup$
    – Michael E2
    Nov 14 '21 at 17:21
  • $\begingroup$ @MichaelE2 you are right. Thank you very much. $\endgroup$
    – Kheeyal
    Nov 14 '21 at 18:01
3
$\begingroup$

Consider the Taylor series around zero of f[5]:

f[n_] := (-(-1 + Sqrt[1 + y^(2 (n - 2))])^( 1/3) + (1 + Sqrt[1 + y^(2 (n - 2))])^(1/3))^4/ (y^((n - 2)/3)); Series[f[5],{y,0,2}]

enter image description here

You see that the first term goes like 1/y. However the integral of 1/y is Log[y]. And this diverges at y==0.

$\endgroup$
1
  • $\begingroup$ Thank you very much $\endgroup$
    – Kheeyal
    Nov 14 '21 at 18:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.