# How do I find all maxima of this function?

I have to find all the maxima of the function

fun[t_?NumberQ] := NIntegrate[1/y^(11/3) Cos[y t], {y, 1, 2 }]


To find the poit with zero first derivative I have done:

Reduce[1/y^(11/3) Cos[y t] == 0, t]


with result:

C \[Element] Integers &&  y != 0 && (t == (-(\[Pi]/2) + 2 \[Pi] C)/y ||  t == (\[Pi]/2 + 2 \[Pi] C)/y)


Now how do I find which of these points are the maxima? I tought to use the second derivative of fun[t], which is the first derivative of the function inside the integral, but I don't know how to use the result of Reduce

• I think fun'[t] is given by Integrate[D[1/y^(11/3) Cos[y t], t], {y, 1, 2}]. (You can also find the integral for fun symbolically as well, which might make finding the maxima easier.) – Michael E2 Aug 26 at 12:14
• Have you tried NMaximize? It seems the global maximum is at t==0, with fun == -(3/64) (-8 + 2^(1/3)) == 0.315941. You can use FindMaximum for the local maxima (e.g., there next maxima are at $\pm$ 4.976, etc.). – AccidentalFourierTransform Aug 26 at 13:02

Clear["Global*"]

fun[t_] = Integrate[1/y^(11/3) Cos[y t], {y, 1, 2}] // Simplify

(* 1/2 (ExpIntegralE[11/3, -I t] + ExpIntegralE[11/3, I t]) - (
ExpIntegralE[11/3, -2 I t] + ExpIntegralE[11/3, 2 I t])/(8 2^(2/3)) *)


Plot fun to obtain initial estimates for use with FindRoot

plt = Plot[fun[t], {t, -15, 15},
WorkingPrecision -> 15] sol = FindRoot[fun'[t] == 0, {t, #}, WorkingPrecision -> 15] & /@
Range[-10, 10, 5]

(* {{t -> -11.0583535838776}, {t -> -4.97614346331708}, {t -> 0}, {t ->
4.97614346331708}, {t -> 11.0583535838776}} *)


Visually verifying:

Show[plt, Epilog -> {Red, AbsolutePointSize,
Point[{t, fun[t]} /. sol]}] The n-th maxima can be computed using

nthMax[n_Integer] := Quiet[
FindMaximum[
NIntegrate[1/y^(11/3) Cos[y t], {y, 1, 2}]
, {t, -8 + 2 π n}]
, {NIntegrate::inumr}]
nthMax /@ Range // TeXForm
`

$$\left( \begin{array}{cc} 0.315941 & \{t\to -\text{2.189*{}^{\wedge}-16}\} \\ 0.148796 & \{t\to 4.97614\} \\ 0.0821207 & \{t\to 11.0584\} \\ 0.0553822 & \{t\to 17.269\} \\ 0.0415163 & \{t\to 23.5158\} \\ \end{array} \right)$$

If you want an analytic formula for large $$n$$, you can use $$-2 \tan ^{-1}\left(\sqrt{\frac{\sqrt{64+8\ 2^{2/3}}+3 \sqrt{2}}{8+\sqrt{2}}}\right)+2 \pi n+\frac{1+\frac{6}{\sqrt{4+\frac{1}{\sqrt{2}}}}}{(3 \pi ) n}+\mathcal O(1/n^2)$$

Here is the numeric result minus the asyptotic expansion, keeping the $$\mathcal O(n),\mathcal O(1)$$, and $$\mathcal O(1/n)$$ approximations, respectively: 