Integral with unreliable result

Question

I want to calculate $\int_R^1 \sqrt{r} |\cos((k+\frac{1}{2})\pi r)|dr $ and I get a result from Mathematica. Then I try to check the result putting the value of $k$ and $R$, (k=1 and R=0.5) in the result and performing a NIntegrate with the same value and the result is different. In the analytical result if you put $k=10$ and $R=0.5$ the result is negative and of course wrong, but if I use Nintegrate the result is posivite. What happened? I am interested in the following function of $R$ and $k$.

$$\text{Assuming}\left[R>0\textrm{&&}k>1\textrm{&&}R<1\textrm{&&}k\in\textrm{Integers},\int_0^R \sqrt{r} \textrm{Abs}\left[\cos\left(k+\frac{1}{2}\right)\pi r\right] dr\right]$$ and Mathematica gives the answer $$\frac{2 S\left(\sqrt{2 k R+R}\right) \sec \left(\pi k R+\frac{\pi R}{2}\right)-2 \sqrt{2 k R+R} \tan \left(\pi k R+\frac{\pi R}{2}\right)}{\pi (2 k+1)^{3/2} \sqrt{\sec ^2\left(\pi k R+\frac{\pi R}{2}\right)}}$$

and if you evaluate this function in $R=0.2$ and $k=1$ the result is $-0.0488018$. This result is different from the definite integral with the same value for the parameters. Thank you in advance again

Answer 1

I get the same result from NIntegrate and Integrate.

Integrate[Sqrt[r] Cos[(k + 1/2) Pi r], {r, R, 1}, 
  Assumptions -> R > 0] /. {k -> 10, R -> 0.5}
NIntegrate[Sqrt[r] Cos[(10 + 1/2) Pi r], {r, 0.5, 1}]

Output:

0.0459518

0.0459518

Answer 2

I assume that |...| means Abs[...].

Define the symbolic integral.

int[r0_?NumericQ, k_?NumericQ] := Integrate[Sqrt[r] Abs[Cos[(k + 1/2) \[Pi] r]], {r, r0, 1}]

Define the numerical integral.

intN[r0_?NumericQ, k_?NumericQ] := NIntegrate[Sqrt[r] Abs[Cos[(k + 1/2) \[Pi] r]], {r, r0, 1}]

Compare these integrals for your chosen parameter values, using 1/2 rather than 0.5 to force symbolic evaluation in int[1/2, 10].

{intN[##], {N[#], #} &@int[##]} &[1/2, 10]

(* {0.269865, 
  {0.269865,
  (1/(441 \[Pi]))(21 + 12 Sqrt[35] + 4 Sqrt[231] + 4 Sqrt[273] + 
  4 Sqrt[357] + 4 Sqrt[399] - 2 Sqrt[21] FresnelS[Sqrt[21/2]] + 
  4 Sqrt[21] FresnelS[Sqrt[11]] - 4 Sqrt[21] FresnelS[Sqrt[13]] + 
  4 Sqrt[21] FresnelS[Sqrt[15]] - 4 Sqrt[21] FresnelS[Sqrt[17]] + 
  4 Sqrt[21] FresnelS[Sqrt[19]] - 2 Sqrt[21] FresnelS[Sqrt[21]])}} *)

The results are numerically the same, and you've got a symbolic expression to use as well.