How to Show That an Integral-Defined Function Is Strictly Decreasing

If a function is given by an integral that depends on a parameter, its decrease often follows from how the integrand changes with that parameter. There are two standard routes: compare the integrals directly, or differentiate the integral and determine the derivative’s sign.

The exact formula matters: without it, no one can verify that a particular function is strictly decreasing. The criteria below apply to a broad class of integral-defined functions and show what must be checked in a specific problem.

The direct comparison method

Suppose

$$F(x)=\int_E K(x,t)w(t)\,dt,$$

where $x$ is the parameter, $t$ is the integration variable, and $w(t)\geq 0$. To show strict decrease, take any $x_1<x_2$ in the domain and compare:

$$F(x_1)-F(x_2)=\int_E [K(x_1,t)-K(x_2,t)]w(t)\,dt.$$

If $K(x_1,t)\geq K(x_2,t)$ for almost every $t$, the difference is nonnegative, so $F(x_1)\geq F(x_2)$. For strict decrease, the integrand in this difference must be positive on a set of positive measure where the weight is also positive. Positivity at just one isolated value of $t$ is not enough: changing an integrand at a single point does not change an ordinary integral.

This approach is often the cleanest when the parameter dependence is visibly monotone. It also avoids differentiating under the integral sign.

The derivative method

If you can differentiate under the integral sign, then

$$F'(x)=\int_E \frac{\partial K}{\partial x}(x,t)w(t)\,dt.$$

If this derivative exists and is negative for every $x$ in the interval, the mean value theorem implies that $F$ is strictly decreasing there. In practice, it is enough to show that $\partial K/\partial x\leq 0$ almost everywhere and is strictly negative on a set of positive measure where $w>0$, for each relevant $x$.

The interchange of differentiation and integration needs justification. A convenient sufficient condition on a finite integration interval is that $K$ and its partial derivative with respect to $x$ are continuous in the variables on the region under consideration. On an infinite interval, a common sufficient condition is an integrable bound for $|\partial K/\partial x|$ that works uniformly for $x$ in a neighborhood of the point where you differentiate. More general results are available, but the assumptions of the applicable theorem must be checked.

Worked example

Consider

$$F(x)=\int_0^1 e^{-xt}\,dt,\qquad x\geq 0.$$

For $x_1<x_2$ and every $t>0$, we have $e^{-x_1t}>e^{-x_2t}$. The inequality holds throughout the interval $(0,1]$, which has positive measure. Integrating gives

$$F(x_1)>F(x_2).$$

Therefore $F$ is strictly decreasing on $[0,\infty)$. The value at $t=0$ does not affect strictness: the integrands agree there, but differ on the rest of the integration interval.

You can also verify the result by differentiation. Because the integrand and its derivative with respect to $x$ are continuous on any bounded rectangle in $(x,t)$, differentiation under this finite-interval integral is valid:

$$F'(x)=-\int_0^1 t e^{-xt}\,dt<0.$$

The integrand $t e^{-xt}$ is positive for $0<t\leq 1$, so the integral is positive and the minus sign makes $F'(x)$ negative. This confirms strict decrease.

For $x>0$, direct integration also gives $F(x)=(1-e^{-x})/x$, while $F(0)=1$. The integral representation proves monotonicity without needing to differentiate that quotient or handle its apparent indeterminate form at zero.

What if the derivative is only nonpositive?

A derivative satisfying $F'(x)\leq 0$ generally proves that $F$ is nonincreasing, not necessarily strictly decreasing. For example, a function can be constant on part of its domain and decrease elsewhere. To establish strict decrease from a derivative argument, prove $F'(x)<0$ throughout the interval, or use another argument that rules out flat portions.

Likewise, if the kernel is only nonincreasing in $x$, the direct comparison gives a nonincreasing integral. You need strict inequality on a set of positive measure to conclude strict decrease.

Common mistakes

• Ignoring the weight’s sign: If $w$ can be negative, the sign of the kernel difference alone may not determine the sign of the integral.
• Using pointwise strictness without checking measure: Strict inequality at one point, or on a measure-zero set, does not make the integral inequality strict.
• Differentiating without justification: The formal derivative of an integrand need not equal the derivative of its integral unless suitable conditions allow the interchange.
• Checking only one pair of inputs: Strict decrease requires $F(x_1)>F(x_2)$ for every pair $x_1<x_2$, or a theorem such as the mean value theorem applied to a negative derivative across the domain.
• Confusing decreasing with strictly decreasing: A nonpositive derivative or a nonincreasing kernel may allow equality for some distinct inputs.

When applying these methods to a special-function representation, identify the parameter, the kernel, and any weight first. Then check the sign and strictness conditions—and state the regularity assumptions that justify the integral comparison or differentiation.