Reflection Identity For Integrals

Why Reflect an Integral?

A definite integral records the total signed accumulation of a function across an interval. Sometimes the integrand is difficult to evaluate directly, but it becomes simpler when we compare its value at \(x\) with its value at the point reflected across the midpoint of the interval.

On \([a,b]\), the reflection sends \(x\) to \(a+b-x\). It exchanges the endpoints: \(a\) reflects to \(b\), and \(b\) reflects to \(a\). Although the points are reversed, the integral of a function over the interval is unchanged when the function is replaced by its reflection. This gives a way to pair expressions without changing the value of the integral.

Definition: For an interval \([a,b]\), the reflection of \(x\) across the midpoint is \(R(x)=a+b-x\). The reflection identity states that, for an integrable function \(f\) on \([a,b]\), \[ \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx. \]

Deriving the Reflection Identity

Assume \(a<b\) and \(f\) is continuous on \([a,b]\). Continuity is a sufficient condition for the definite integrals below to exist and for the change of variable to apply. In the integral on the right, set \(u=a+b-x\). Then \(x=a+b-u\), so \(dx=-du\). When \(x=a\), \(u=b\); when \(x=b\), \(u=a\). Both the minus sign and the reversed limits matter:

\begin{aligned} \int_a^b f(a+b-x)\,dx &=\int_b^a f(u)(-du)\\ &=\int_a^b f(u)\,du\\ &=\int_a^b f(x)\,dx. \end{aligned}

The last equality just renames the variable of integration: \(u\) and \(x\) are both dummy variables. The calculation proves the identity.

Theorem: If \(f\) is integrable on \([a,b]\), then \[ \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx. \] In particular, this holds whenever \(f\) is continuous on \([a,b]\).

The map \(R(x)=a+b-x\) is an involution: reflecting twice returns the original point, since \(R(R(x))=x\). It maps the interval onto itself but reverses its direction. The change of variable proof accounts for both features. Thus the identity is about the value of the integral over the whole interval; it does not assert that \(f(x)=f(a+b-x)\) at each point.

Pairing an Integrand with Its Reflection

Since the original integrand and its reflection have the same integral, we may add or average them. For example, if \(I=\int_a^b f(x)\,dx\), then

2I=\int_a^b \bigl(f(x)+f(a+b-x)\bigr)\,dx.

This follows by adding two equal integrals and using linearity of the definite integral. The paired integrand can be simpler than either term alone. The related difference also has a useful property:

\int_a^b \bigl(f(x)-f(a+b-x)\bigr)\,dx=0.

Indeed, the two terms have equal integrals, so their difference integrates to zero. This conclusion concerns the integral of the difference; it does not mean that the difference is zero at every point.

Worked Example: A Rational Function on the Unit Interval

Evaluate \[ I=\int_0^1 \frac{x^2}{x^2+(1-x)^2}\,dx. \] The denominator is unchanged by replacing \(x\) with \(1-x\), while the numerator changes. Apply the reflection identity on \([0,1]\), where \(a+b=1\):

I=\int_0^1 \frac{(1-x)^2}{(1-x)^2+x^2}\,dx.

Add this expression for \(I\) to the original expression. The denominators match, and the numerators add to the denominator:

\begin{aligned} 2I &=\int_0^1 \left( \frac{x^2}{x^2+(1-x)^2} + \frac{(1-x)^2}{x^2+(1-x)^2} \right)\,dx\\ &=\int_0^1 \frac{x^2+(1-x)^2}{x^2+(1-x)^2}\,dx\\ &=\int_0^1 1\,dx\\ &=1-0=1. \end{aligned}

Therefore \(I=\frac12\). To check the result, the reflected integral has the same value as \(I\), and their integrands add to \(1\). Their two integrals must therefore add to \(\int_0^1 1\,dx=1\), consistent with \(I=\frac12\).

Reflection on an Interval Other Than \([0,1]\)

The reflected point is determined by the two endpoints, not by a special choice of interval. On \([a,b]\), the reflection is \(a+b-x\), so on \([0,\pi]\) it is \(\pi-x\), and on \([1,3]\) it is \(4-x\). The same identity applies on each interval as long as the integrand is integrable there.

Worked Example: A Trigonometric Integral

Evaluate \[ I=\int_0^\pi x\sin x\,dx. \] Reflect \(x\) to \(\pi-x\). Since \(\sin(\pi-x)=\sin x\), the reflection identity gives

I=\int_0^\pi (\pi-x)\sin(\pi-x)\,dx =\int_0^\pi (\pi-x)\sin x\,dx.

Add the two equal expressions for \(I\):

\begin{aligned} 2I &=\int_0^\pi \bigl(x\sin x+(\pi-x)\sin x\bigr)\,dx\\ &=\int_0^\pi \pi\sin x\,dx\\ &=\pi\bigl[-\cos x\bigr]_0^\pi\\ &=\pi\bigl(-\cos\pi-(-\cos 0)\bigr)\\ &=\pi\bigl(1-(-1)\bigr)=2\pi. \end{aligned}

Hence \(I=\pi\). Check independently by integration by parts, taking \(u=x\) and \(dv=\sin x\,dx\), so \(du=dx\) and \(v=-\cos x\):

\begin{aligned} \int_0^\pi x\sin x\,dx &=\bigl[-x\cos x\bigr]_0^\pi+\int_0^\pi \cos x\,dx\\ &=-\pi\cos\pi-0+\bigl[\sin x\bigr]_0^\pi\\ &=\pi+0=\pi. \end{aligned}

The direct calculation agrees with the result from reflection.

Using Reflection to Detect Cancellation

If reflecting an expression changes its sign, the integral of that expression must be zero. More precisely, suppose \(g(a+b-x)=-g(x)\) for every \(x\) in \([a,b]\). The reflection identity then gives \(\int_a^b g(x)\,dx=\int_a^b -g(x)\,dx=-\int_a^b g(x)\,dx\). Therefore the integral equals zero. This is a consequence of the identity and linearity, not an additional assumption about the size or sign of the function.

Worked Example: A Logarithmic Difference

Evaluate \[ J=\int_0^1 \ln\left(\frac{1+x}{2-x}\right)\,dx. \] The logarithm is defined and continuous throughout \([0,1]\), since both \(1+x\) and \(2-x\) are positive there. Reflect \(x\) to \(1-x\):

\begin{aligned} \ln\left(\frac{1+(1-x)}{2-(1-x)}\right) &=\ln\left(\frac{2-x}{1+x}\right)\\ &=-\ln\left(\frac{1+x}{2-x}\right). \end{aligned}

The reflected integrand is the negative of the original one, so

J=\int_0^1 -\ln\left(\frac{1+x}{2-x}\right)\,dx=-J.

Thus \(2J=0\), and \(J=0\). For a direct check, split the logarithm into a difference and evaluate each integral:

\begin{aligned} J &=\int_0^1 \ln(1+x)\,dx-\int_0^1 \ln(2-x)\,dx\\ &=(2\ln 2-1)-(2\ln 2-1)\\ &=0. \end{aligned}

For the first integral, an antiderivative is \((1+x)\ln(1+x)-(1+x)\), giving \(2\ln 2-1\) between \(0\) and \(1\). For the second, set \(u=2-x\); its limits become \(2\) and \(1\), and \(dx=-du\), so it has the same value. The direct calculation confirms the reflection argument.

A Common Sign Error

When substituting \(u=a+b-x\), the new limits occur in the opposite order. A frequent mistake is to reverse the limits but forget the minus sign from \(dx=-du\). That incorrectly changes the sign of the integral.

Common mistake: Writing \[ \int_a^b f(a+b-x)\,dx=\int_b^a f(u)\,du \] without accounting for \(dx=-du\). The correct substitution is \[ \int_a^b f(a+b-x)\,dx =\int_b^a f(u)(-du) =\int_a^b f(u)\,du. \]

The sign is not optional: reversing the limits changes the sign of an integral, and the negative differential changes it back. As a quick check, take \(f(x)=1\). The original integral is \(b-a>0\). The reflected integrand is still \(1\), so its integral must also be \(b-a\), not \(a-b\).

When the Identity Is Useful

Reflection is most useful when the reflected expression combines cleanly with the original. Before evaluating an integral directly, inspect the interval endpoints and ask what \(a+b-x\) becomes. Then compare the two expressions. Common helpful outcomes include a constant sum, a sign reversal, or a simplification of factors that were paired around the midpoint.

Use the identity only on the same interval, with the reflected input \(a+b-x\).
Check that the integrand is integrable on the interval; continuity is a convenient sufficient condition.
When changing variables, reverse the limits and include the negative differential.
Do not conclude that the original and reflected integrands are pointwise equal merely because their integrals are equal.

The reflection identity does not by itself evaluate every integral. Its value is that it gives an equal integral in a second form. The main step is to notice whether that second form creates a useful pair.

Check Your Understanding

Use the reflection identity and show the key steps in each answer.

What is the reflection of \(x\) on the interval \([2,5]\)?
If \(I=\int_0^1 \frac{x}{x+(1-x)}\,dx\), what does reflection tell you about the reflected integrand, and what is \(I\)?
Suppose \(g(3-x)=-g(x)\) on \([1,2]\). What is \(\int_1^2 g(x)\,dx\)?
In the substitution \(u=a+b-x\), what happens to the limits and to the differential?
Does \(\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx\) imply \(f(x)=f(a+b-x)\) at every point? Explain briefly.

Reflection Identity For Integrals

What You'll Learn

Why Reflect an Integral?

Deriving the Reflection Identity

Pairing an Integrand with Its Reflection

Worked Example: A Rational Function on the Unit Interval