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I have a question. Is this integral improper? $$\int_0^\infty \frac{5x}{e^x-e^{-x}} \, dx = \int_0^a \frac{5x}{e^{x}-e^{-x}} \, dx+ \int_a^\infty \frac{5x}{e^x-e^{-x}} \, dx$$

Why is $\displaystyle\int_0^a \frac{5x}{e^{x}-e^{-x}}dx$ an improper integral at point $x=0$? And $\displaystyle\int_a^\infty \frac{5x}{e^{x}-e^{-x}} \, dx$ is improper integral at $x \to \infty $?


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It's improper at $x=0$ because the function isn't defined there. It's improper at $x=\infty$ also because the function isn't defined at $\infty$. – Git Gud Feb 17 '13 at 17:10
@GitGud : I don't think that's really the essence of the matter. – Michael Hardy Feb 17 '13 at 17:15
@MichaelHardy Me neither, that's why I just left it as a comment. – Git Gud Feb 17 '13 at 17:17
@Git Gud Thanks, but why function isn't defined? In first one I have $(0/0)$ and by L'hopital rule: $\lim_{x \to 0 } \frac{5x}{e^{x}-e^{-x}}=\lim_{x \to 0 }\frac{5}{e^{x}+e^{-x}}=\frac{5}{2}$ – Panka Feb 17 '13 at 17:24
For those interested in evaluating the integral, we have that $$\int_{0}^{\infty}\frac{x}{e^{x}+e^{-x}}dx=\int_{0}^{\infty}\frac{x}{e^{x}} \left(\frac{1}{1+e^{-2 x}}\right)dx $$ which equals $$\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{\infty}xe^{-(2k+1)x}dx.$$ Letting $u=(2k+1)x$, the above becomes $$\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}\int_{0}^{\infty}ue^{-u}du,$$ and since $\Gamma(2)=\int_{0}^{\infty}ue^{-u}du,$ we have that $$\int_{0}^{\infty}\frac{x}{e^{x}+e^{-x}}dx=C$$ where $C$ is Catalan's Constant – Eric Naslund Feb 17 '13 at 17:28
up vote 3 down vote accepted

An integral $$ \int_a^b f(x) \, dx $$ is improper at $a$ if it must be defined as $$ \lim_{c\downarrow a} \int_c^b f(x)\,dx. $$ In Lebesgue's theory of integration, that wuold be necessary only if for every $c>a$, the integral $\displaystyle\int_a^c f(x)\,dx$ cannot be defined because the integrals of the positive and negative parts are both infinite. For example $\displaystyle \int_0^\infty \frac{\sin x}{x}\,dx$ is improper at $\infty$ because if one integrates only over those regions where $(\sin x)/x$ is positive, one gets $+\infty$, and if one integrates only over regions where $(\sin x)/x$ is negative, one gets $-\infty$. But $\displaystyle\lim_{b\to\infty} \int_0^b \frac{\sin x}{x}\,dx$ nonetheless exists and is finite.

This doesn't happen with $\displaystyle\int_0^b \frac{5x}{e^x-e^{-x}}\,dx$ at $0$. But in other contexts than Lebesgue's theory, when it happens that the only good way to find $\displaystyle\int_0^b$ is by finding $\displaystyle\lim_{c\downarrow 0}\int_c^b$, then one might call it an improper integral. In this case, the function approaches $5$, and thus remains bounded, as $x$ approaches $0$, and that's how one knows it's not improper in Lebesgue's sense. But the function is undefined at $0$, and therefore so is its antiderivative, so if you're using the fundamental theorem of calculus, i.e. if you're finding the integral by finding the antiderivative, it may be that taking the limit is the only way to find the integral, and so in that sense it's improper.

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Maybe you can try a Taylor series expansion. Make the lower limit of your first integral on the RHS $L$ such that $0 < L \ll 1$, then let $L < a \ll 1$. You can do a Taylor series expansion of the denominator which will be $2x + O(x^3)$. You can cancel a factor of $x$ giving the integrand $(5/2)\cdot(1+O(x^2))^{-1}$. Using the Taylor series of $\frac{1}{1+x}$, your integral will be $\frac{5x}{2} + O(x^3)$. Take the limit as $L$ goes to zero.

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