I have a function like this.. $$ f[k]=\int_0^\infty Exp[-ax]x(Cos[bx])^{2}BesslJ[0,kx]dx $$ I want to know how many local maximuns this function has.. So I try to get the derivative of f[k] $$ g[k]=D[f[k],k]; $$ Then the result from mathematica is $$ g[k]=-6ka^{-4}(1+k^{2}/a^{2})^{-5/2}-3k(a-2ib)^{-4}(1+k^{2}/(a-2ib)^{2})^{-5/2}-3k(a+2ib)^{-4}(1+k^{2}/(a+2ib)^{2})^{-5/2},. $$

I need to get the results $$ g[k]==0,k=?? $$ Because in the function there is a & b,I use Solve..But it can't get the answer..Actually just the number of the solutions is necessary for me.. I am a beginner of the Mathematica..TKS in advance.

  • 2
    $\begingroup$ Please post the code for f[x]. LaTeX is fine for typesetting, but it's better to have actual code. $\endgroup$ – DumpsterDoofus Feb 4 '15 at 0:51
  • $\begingroup$ TKS for your suggestion.. $\endgroup$ – Cici Feb 4 '15 at 0:57
  • $\begingroup$ One solution is x = 0. I believe that there are two more. $\endgroup$ – bbgodfrey Feb 4 '15 at 1:07
  • $\begingroup$ YES..x=0 is one of the answer...But I need the exact numbers of this function...Any solutions? $\endgroup$ – Cici Feb 4 '15 at 1:09
  • $\begingroup$ Are a and b both real with a > 0? $\endgroup$ – bbgodfrey Feb 4 '15 at 1:54

g[k]=0 is not easily simplified. So, I plotted g (with k and b both scaled to a):

Manipulate[ Plot[g /. {a -> 1, b -> b0}, {k, 0, 20}, PlotRange -> {-.01, .01}],
 {{b0, 1}, 0, 10, Appearance -> "Labeled"}]

A sample curve, for b = 6.12, is

g at 6.12

Manifestly, there are three zeros, one at k = 0 and the others at positions dependent on b. (The curve does not cross the axis at large k.)


It also is instructive to plot f itself (again for b = 6.12)

f at 6.12

Thus, f has 2 maxima and one minimum.

| improve this answer | |
  • $\begingroup$ YES..U R right..I plan to try Manipulate k&a&b before but mathematica can't give the answer..U assumed a→1(or maybe some others) and now it works. TKS so much. $\endgroup$ – Cici Feb 4 '15 at 3:16
  • $\begingroup$ Actually a->1 is a renormalization without loss of generality. Glad to be of help. $\endgroup$ – bbgodfrey Feb 4 '15 at 3:17

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