Changing the Finite Endpoint Without Changing Tail Convergence

Why the Finite Starting Point Is Not the Whole Story

An improper integral over an infinite interval asks whether the accumulated integral settles to a finite value as the upper limit grows without bound. It is natural to wonder whether that answer depends on the chosen finite starting point. If a function is well behaved between two possible starting points, changing from one to the other only adds or removes a finite piece of area. That finite change can affect the value, but not whether the tail converges.

This distinction is useful whenever an integral is easier to analyze from one finite point than another. The key requirement is that the function have a finite definite integral on the interval between those points. The argument does not require $f$ to be positive.

Definition and Endpoint-Shift Formula

Suppose $f$ is integrable on every closed bounded interval contained in $[a,\infty)$. In particular, this condition holds if $f$ is continuous there. The improper integral from $a$ to infinity is defined by a limit, when that finite limit exists.

\int_a^\infty f(x)\,dx = \lim_{R\to\infty}\int_a^R f(x)\,dx.

Here, convergence means that the limit exists as a finite real number. The same definition applies from any other finite starting point $b$ in the domain. For finite endpoints, use the oriented-integral convention $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$.

Theorem: Suppose $f$ is integrable on every closed bounded interval between finite points $a$ and $b$, and on every bounded interval from either point onward. Then $\int_a^\infty f(x)\,dx$ converges if and only if $\int_b^\infty f(x)\,dx$ converges. When they converge, $$ \int_b^\infty f(x)\,dx = \int_a^\infty f(x)\,dx-\int_a^b f(x)\,dx. $$

The oriented integral handles either ordering of $a$ and $b$. If $b>a$, moving the starting point to the right removes the finite piece from $a$ to $b$. If $b<a$, the oriented integral $\int_a^b f(x)\,dx$ is negative, so subtracting it adds the piece from $b$ to $a$.

Why the Theorem Works

The proof rests on additivity of the definite integral over finite intervals. For every $R$ greater than both $a$ and $b$, that property gives an exact identity. The identity does not depend on taking a limit.

Split at the other endpoint.
For $R>\max(a,b)$, additivity gives $\int_a^R f(x)\,dx=\int_a^b f(x)\,dx+\int_b^R f(x)\,dx$.

Rearrange the finite identity.
Since $\int_a^b f(x)\,dx$ is a finite constant, $\int_b^R f(x)\,dx=\int_a^R f(x)\,dx-\int_a^b f(x)\,dx$.

Take the limit.
If $\int_a^R f(x)\,dx$ has a finite limit as $R\to\infty$, subtracting the fixed finite constant leaves a finite limit. That limit is the stated formula.

This proves that convergence from $a$ implies convergence from $b$. The same identity can be rearranged as $\int_a^R f(x)\,dx=\int_b^R f(x)\,dx+\int_a^b f(x)\,dx$. Thus, if the integral from $b$ converges, the integral from $a$ converges as well. Both directions are established.

The finite-interval integrability condition matters. Without it, the term $\int_a^b f(x)\,dx$ might not be a finite number, and the proof would not establish a finite constant shift. This theorem concerns changing a starting point across an interval where the ordinary definite integral exists; it does not analyze singularities within that interval.

Worked Examples

Worked Example: Moving the Starting Point to the Right

Let $f(x)=\frac{1}{(x+1)^2}$, and compare the integrals starting at $0$ and $2$. The function is continuous for $x\geq 0$, so the finite interval from $0$ to $2$ has a definite integral.

An antiderivative is $F(x)=-\frac{1}{x+1}$, since differentiating gives $F'(x)=\frac{1}{(x+1)^2}$. First evaluate the integral from $0$ to infinity:

\int_0^\infty \frac{1}{(x+1)^2}\,dx = \lim_{R\to\infty}\left[-\frac{1}{x+1}\right]_0^R = \lim_{R\to\infty}\left(1-\frac{1}{R+1}\right) =1.

Now calculate the finite piece from $0$ to $2$:

\int_0^2 \frac{1}{(x+1)^2}\,dx = \left[-\frac{1}{x+1}\right]_0^2 = -\frac13-(-1) = \frac23.

The endpoint-shift formula therefore gives:

\int_2^\infty \frac{1}{(x+1)^2}\,dx = 1-\frac23 = \frac13.

Check directly using the same antiderivative: $\lim_{R\to\infty}\left[-\frac{1}{x+1}\right]_2^R=\lim_{R\to\infty}\left(\frac13-\frac{1}{R+1}\right)=\frac13$. Differentiating $F$ gives the original integrand, and the direct limit confirms the shifted value.

Worked Example: Moving the Starting Point to the Left

Let $f(x)=e^{-x}$, and compare starting points $3$ and $1$. Since $1<3$, the integral from $3$ to $1$ in the shift formula is oriented backwards. An antiderivative is $F(x)=-e^{-x}$, because $F'(x)=e^{-x}$.

Starting at $3$, evaluate the improper integral:

\int_3^\infty e^{-x}\,dx = \lim_{R\to\infty}\left[-e^{-x}\right]_3^R = \lim_{R\to\infty}(e^{-3}-e^{-R}) = e^{-3}.

The finite integral in the shift formula is:

\int_3^1 e^{-x}\,dx = \left[-e^{-x}\right]_3^1 = -e^{-1}-(-e^{-3}) = e^{-3}-e^{-1}.

Consequently, the integral starting at $1$ is:

\int_1^\infty e^{-x}\,dx = \int_3^\infty e^{-x}\,dx-\int_3^1 e^{-x}\,dx = e^{-3}-(e^{-3}-e^{-1}) = e^{-1}.

For a direct check, $\lim_{R\to\infty}[-e^{-x}]_1^R=\lim_{R\to\infty}(e^{-1}-e^{-R})=e^{-1}$. The derivative of the antiderivative is $e^{-x}$, and the direct limit agrees with the endpoint-shift calculation. The negative value of $\int_3^1 e^{-x}\,dx$ is exactly what makes the formula work when moving left.

Worked Example: A Finite Shift Does Not Cure Divergence

Let $f(x)=\frac1x$ for $x\geq1$, and compare starting points $1$ and $4$. The function is continuous on this domain, so $\int_1^4 \frac1x\,dx$ is finite. An antiderivative is $F(x)=\ln x$, whose derivative is $1/x$.

For $R>4$, the two truncated integrals are:

\int_1^R \frac1x\,dx=\ln R-\ln 1=\ln R, \qquad \int_4^R \frac1x\,dx=\ln R-\ln 4.

As $R\to\infty$, $\ln R\to+\infty$, and $\ln R-\ln4\to+\infty$ as well. Neither improper integral has a finite limit, so both diverge. Their truncated values differ by the fixed finite amount $\ln4$: indeed, $\int_1^4 \frac1x\,dx=\ln4-\ln1=\ln4$, and $\int_4^R \frac1x\,dx=\int_1^R \frac1x\,dx-\ln4$. This verifies the finite-interval identity and shows that subtracting a finite constant does not create convergence.

A Common Sign Error

A frequent mistake is to add the finite piece when moving the starting point to the right. For the first example, this would give the incorrect calculation:

\int_2^\infty \frac{1}{(x+1)^2}\,dx \overset{\text{wrong}}{=} \int_0^\infty \frac{1}{(x+1)^2}\,dx + \int_0^2 \frac{1}{(x+1)^2}\,dx = 1+\frac23 = \frac53.

This adds the area that should have been removed. Splitting the integral at $2$ gives the correct identity, $\int_0^\infty f(x)\,dx=\int_0^2 f(x)\,dx+\int_2^\infty f(x)\,dx$. Therefore the correction is to subtract the finite piece:

\int_2^\infty \frac{1}{(x+1)^2}\,dx = 1-\frac23 = \frac13.

The direct check in the first worked example confirms this value. A reliable approach is to write the finite-interval splitting identity before deciding whether the finite piece is added or subtracted.

What Changes—and What Does Not

Changing a finite endpoint changes the numerical value by the definite integral across the interval between the two endpoints. It does not change whether the integral to infinity has a finite limit, provided that intervening finite integral exists. This holds for positive, negative, or sign-changing functions: the proof uses additivity and a finite constant, not positivity.

The result is about convergence, not about whether two integrals have equal values. In general, they do not. It also concerns ordinary convergence of the improper integral as a finite limit. The theorem supplies no conclusion if the function fails to be integrable between the proposed starting points. Keeping these distinctions clear makes endpoint changes a simple accounting step rather than a new test of the infinite tail.

Check Your Understanding

Use the endpoint-shift identity and the definition of an improper integral to answer the following.

State the finite-interval condition needed to compare integrals starting at $a$ and $b$.
If $\int_a^\infty f(x)\,dx$ converges and $\int_a^b f(x)\,dx$ is finite, write the formula for $\int_b^\infty f(x)\,dx$.
Why does adding or subtracting a finite constant preserve the existence of a finite limit?
For $f(x)=e^{-x}$, moving the starting point from $3$ to $1$ uses $\int_3^1 e^{-x}\,dx$. Why is that finite integral negative?
If the truncated integrals from one starting point tend to $+\infty$, what does the endpoint-shift identity imply about another starting point across a finite integrable interval?