Quote from Essential Calculus: Early Transcendentals, by James Stewart:

If $f$ is continuous, then $$\int_{-\infty}^\infty f(x)dx=\lim_{t \to \infty}\int_{-t}^tf(x)dx$$

I thought this would be true if such a limit exists (aka if the area is convergent from $a \to \infty$ and from $-\infty \to a$), but the book answer-sheet marks it as false. Could anyone explain to me why it is false?


  • $\begingroup$ You need to quantify over $f$. $\endgroup$ – Git Gud Mar 13 '14 at 16:35
  • $\begingroup$ @GitGud mind to elaborate? (I forgot to state that $f$ is continuous) $\endgroup$ – Aegis Mar 13 '14 at 16:36

Take for example $f(x)=x$. Then $$ \int_{-t}^t f(x)\,dx=0, $$ since $f$ is an odd function, and hence the limit $\lim_{t\to\infty}\int_{-t}^t f(x)\,dx$ exists and it is equal to zero. However, the function $f(x)=x$ is NOT integrable on the whole of $\mathbb R$.

Another example is $$ \lim_{t\to\infty}\int_{-t}^t \frac{\sin x}{x}\,dx=\pi, $$ although $\dfrac{\sin x}{x}$ is also NOT integrable on the whole of $\mathbb R$.

  • $\begingroup$ Thanks! That made things much clearer. One last thing: would the statement be true if you modified it by saying "if $f$ is integrable on the whole of R then ..." ? $\endgroup$ – Aegis Mar 13 '14 at 16:41
  • $\begingroup$ Yes this is true. Unfortunately, it is not very simple to prove, as we need to use Lebesgue Dominated Convergence Theorem! $\endgroup$ – Yiorgos S. Smyrlis Mar 13 '14 at 16:42
  • $\begingroup$ I might do some reading on that. Thanks a lot! $\endgroup$ – Aegis Mar 13 '14 at 16:43
  • $\begingroup$ The second example is perhaps not completely relevant, since you are talking about Lebesgue integrability (I guess), while the question seems to be about improper Riemann integrals. +1 for the first example, though. $\endgroup$ – Hans Lundmark Mar 13 '14 at 17:58

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