# Improper Integral Exists Since a Limit Exists

I have read a solution which I didn't understand. $$\mbox{Given the improper integral:}\quad \int_{0}^{1}{\,\mathrm{e}^x - \,\mathrm{e}^{-x} - 2x \over 2x^{2}\left(\,\mathrm{e}^x - \,\mathrm{e}^{-x}\,\right)}\,\,\mathrm{d}x$$

The integrand is undefined at $x = 0$, but the lecturer says that since $\displaystyle{% \lim_{x \to 0^{+}}\,{\,\mathrm{e}^{x} - \,\mathrm{e}^{-x} - 2x \over 2x^{2}\left(\,\mathrm{e}^{x} - \,\mathrm{e}^{-x}\,\,\right)}}$ exists,

therefore $x = 0$ is not a singular point, and therefore the improper integral exists, and it is finite. I don't understand why showing that the integrand has a limit means that the definite integral has a limit when approaching $0$.

Thank you.

• If the limit of the integrand at the border exists, it is continuous on a compact interval and thus integrable. – Redundant Aunt Jul 13 '16 at 12:44
• I know the relation of "If a function is continuous then it is integrable" but doesn't that theorem requires the function to be both defined and continuous? – Taru Jul 13 '16 at 12:46

One of the big facts of integrability is: If $f$ is continuous on $[a,b]$, then $f$ is integrable on $[a,b]$.

If $f$ is continuous on $(a,b]$, and $\lim_{x\to a^+} f(x) = L$, then we can create a new function $g$ on $[a,b]$ by $$g(x) = \begin{cases} f(x) & a < x \leq b \\ L & x=a \end{cases}$$ Then by design, $g$ is continuous on $[a,b]$.

We claim that the improper integral $\int_a^b f(x)\,dx$ converges to $\int_a^b g(x)\,dx$. Because, $$\lim_{x\to a^+} \int_x^b f(t)\,dt = \lim_{x\to a^+} \int_x^b g(t)\,dt$$ since $f(t) = g(t)$ for all $t$ in $(a,b]$. Also, since $g$ is integrable, the function $x\mapsto \int_x^b g(t)\,dt$ is continuous on $[a,b]$. So $$\lim_{x\to a^+} \int_x^b g(t)\,dt = \int_a^b g(t)\,dt$$ This establishes the claim.

• Your answer had really helped me understand a few more things on both integration and limits. Thank you very much. – Taru Jul 13 '16 at 13:23
• @Taru: You're welcome. It's nice to have a question on here that isn't just someone's homework problem. – Matthew Leingang Jul 13 '16 at 13:43

I can't claim that analogous term exists in English, but in my mother tongue, we can talk of so called "removable discontinuities", which can be removed by defining the function with its limit in discontinuity. Having done that you can treat it like an ordinary continuous function. Since every continuous on [a; b] function is integrable, the integral exists (you just think that you integrate new continuous function).

• Indeed, that phenomenon is called removable discontinuity in English, too. – Matthew Leingang Jul 13 '16 at 12:51
• Try to build the plot with a tool like WolframAlpha. The machine won't even notice the discontinuity. – Konstantin K Jul 13 '16 at 12:53
• @ Matthew Leingang Thanks, Matthew – Konstantin K Jul 13 '16 at 12:55

While defining integrals such as the one you have, it is customary to write

\begin{eqnarray} \int_{0}^{1}\frac{e^{x}-e^{-x}-2x}{2x^{2}(e^{x}-e^{-x})}~dx=\lim\limits_{t\rightarrow 0^{+}}~\int_{t}^{1}\frac{e^{x}-e^{-x}-2x}{2x^{2}(e^{x}-e^{-x})}~dx. \end{eqnarray}

Equivalently, in writing the given integral as a limit, we are asking for the integrand to be well defined for all $x\in (0,1]$.

A sufficient condition to ensure this is that $\lim\limits_{x\rightarrow 0^{+}} \frac{e^{x}-e^{-x}-2x}{2x^{2}(e^{x}-e^{-x})}$ exists.

The integrand function can be written as $$f(x)=\frac{\sinh(x)-x}{2x^2\sinh(x)}$$ that is continuous over $(0,1]$ and bounded in a right neighbourhood of the origin, since $$\sinh(x) = x+\frac{x^3}{6}+o(x^4)$$ gives $\lim_{x\to 0^+}f(x) = \color{red}{\frac{1}{12}}$. Integrability readily follows.
By computing the Taylor series of $f(x)$ we also have: $$\forall x\in[0,1],\qquad 0\leq f(x)\leq \frac{1}{12}-\frac{7 x^2}{720}+\frac{31 x^4}{30240},$$ hence it follows that: $$0\leq \int_{0}^{1}f(x)\,dx \leq \frac{1349}{16800}.$$

In fact if $f$ is continuous and bounded on $(0,1],$ then $\lim_{a\to 0^+}\int_a^1 f(x)\, dx$ exists, i.e., the improper integral $\int_0^1 f(x)\,dx$ converges. Thus improper integrals such as $\int_0^1\sin (1/x)\,dx$ converge. (The case where $f(x)$ is continuous on $(0,1]$ and has a finite limit as $x\to 0^+,$ such as in your specific problem, is contained in this result.)