# Testing Expression over region of values

I'd like to test whether this expression is negative when a and b are constrained to lie within (0,1) interval. Any ideas how?

D[Integrate[((1 - (a*x))^12)*((1 + (a/b))/((1 + (a*x/b))^2)), {x, 0, 1}], a]


Clear["Global*"]

expr1 = ((1 - (a*x))^12)*((1 + (a/b))/((1 + (a*x/b))^2)) // Simplify;


Evaluating the integral

int = Assuming[0 < a <= 1 && 0 < b <= 1,
Integrate[expr1, {x, 0, 1}] // Simplify];


Taking the derivative

expr2 = D[int, a] // Simplify[#, 0 < a <= 1 && 0 < b <= 1] &;


Numerically finding the minimum

NMinimize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b},
WorkingPrecision -> 20] // N

{-6., {a -> 8.10042*10^-9, b -> 0.91264}}

Limit[expr2, a -> 0]

(* -6 *)


The minimum occurs when a is near 0

Limit[expr2, a -> 0]

(* -6 *)


EDIT: numerically finding the maximum

NMaximize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b},
WorkingPrecision -> 20] // N

(* {-1.00973*10^-12, {a -> 0.981099, b -> 9.7192*10^-13}} *)


The maximum occurs for b near 0

Limit[expr2, b -> 0]

(* 0 *)


Consequently, expr2 is nonpositive everywhere in the region. Graphically,

Plot3D[expr2, {a, 0, 1}, {b, 0, 1},
PlotPoints -> 50,
Exclusions -> True,
WorkingPrecision -> 15,
ClippingStyle -> None]
` • Thanks Bob. Shouldn't the last step be a maximization as I want to verify if the expression is always less than 0 e.g. that it's max value < 0? Also do you know why it isn't possible to do analytically rather than numerically? – David Zentler-Munro Oct 28 at 14:25
• The expression is transcendental. Terms include product of high order polynomial times a log. Analytic solution is unlikely. – Bob Hanlon Oct 28 at 14:45