calculusdefinite integralsintegration
I was trying to find the integral of $\frac{\sin x}{x}$ recently, and nothing I tried appeared to get me any closer to a solution, so I looked it up and apparently it's a Non-integrable function.
What makes a function Non-integrable and is there a way to solve it? Why can't it be solved with the "normal" methods.
Forgive my error-ridden terminology, I'm new.
So, the idea is to split the integrand $\frac{8\sin x+3\cos x}{\sin x+2\cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=\frac{\sin x+2\cos x}{\sin x+2\cos x}$. The other? Well, it would be nice if we could just substitute $\sin x+2\cos x$. That works when the numerator is the derivative $\cos x - 2\sin x$, or a multiple of that.
So, there's our system: $$8\sin x+3\cos x=a(\sin x+2\cos x)+b(-2\sin x+\cos x)$$ Equate coefficients for $$a-2b = 8$$ $$2a+b = 3$$ Can you finish it from there?
I would have guessed this was a duplicate, but I wasn't able to find another instance of this question during a cursory search.
Hint The denominator has period $2 \pi i$, which suggests using the following contour $\Gamma_{\epsilon, R}$, $0 < \epsilon < \pi$, $\epsilon < R$, (for which an illustration was already drawn for an answer to the similar question linked by Zacky in the comments):
The key trick here, which we apply with the benefit of hindsight, is to evaluate instead the similar integral $$\int_{\Gamma_{\epsilon, R}} \frac{z^2 \,dz}{e^z - 1} .$$ The interior of $\Gamma_{\epsilon, R}$ contains no poles, so this integral vanishes. Thus, parameterizing the constituent arcs of the contour gives \begin{multline} 0 = \underbrace{\int_\epsilon^R \frac{x^2 \,dx}{e^x - 1}}_{A} + \underbrace{\int_0^{2 \pi} \frac{(R + i y)^2 \cdot i \,dy}{e^{R + i y} - 1}}_{B} + \underbrace{\int_R^\epsilon \frac{(x + 2 \pi i)^2 \,dx}{e^x - 1}}_{C} \\ + \underbrace{\int_0^{-\pi / 2} \frac{(2 \pi i + \epsilon e^{i\theta})^2 \cdot i \epsilon e^{i \theta} d \theta}{e^{\epsilon e^{i \theta}} - 1}}_{D} + \underbrace{\int_{2 \pi - \epsilon}^\epsilon \frac{(i y)^2 \cdot i \,dy}{e^{i y} - 1}}_{E} + \underbrace{\int_{\pi / 2}^0 \frac{(\epsilon e^{i\theta})^2 \cdot i \epsilon e^{i \theta} d \theta}{e^{\epsilon e^{i \theta}} - 1}}_{F} . \qquad (\ast) \end{multline}
A standard bounding argument shows that $B \to 0$ as $R \to \infty$. Computing the first terms of the Taylor series gives that the integrand of $D$ is $-4 \pi^2 i + O(\epsilon)$, so $D = 2 \pi^3 i + O(\epsilon)$, and similarly $F = O(\epsilon)$ (in fact, the integrand is analytic at $0$, which implies this without any more computation). Now, expanding the integrand of $C$ gives $$-\int_\epsilon^R \frac{x^2 \,dx}{e^x - 1} = -\int_\epsilon^R \frac{x^2 \,dx}{e^x - 1} - 4 \pi i \int_\epsilon^R \frac{x \,dx}{e^x - 1} + 4 \pi^2 \int_\epsilon^R \frac{\,dx}{e^x - 1} .$$ The first term on the r.h.s. cancels $A$, and after taking appropriate limits the second term will be constant multiple of the integral $\color{#df0000}{\int_0^\infty \frac{x \,dx}{e^x - 1}}$ of interest. The third term diverges as $\epsilon \searrow 0$, and it turns out that the diverging part of this term in $\epsilon$ is canceled by the diverging part of $E$, but we can avoid dealing with this issue directly by passing to the imaginary part of $(\ast)$. Computing gives $\operatorname{Im} E = -\frac{1}{2} \int_\epsilon^{2 \pi - \epsilon} y^2 \,dy = -\frac{4}{3} \pi^3 + O(\epsilon)$, so taking the limits $\epsilon \searrow 0, R \to \infty$ of the imaginary part of $(\ast)$ leaves $$0 = -4 \pi \color{#df0000}{\int_0^\infty \frac{x\,dx}{e^x - 1}} + 2 \pi^3 - \frac{4}{3} \pi^3 ,$$ and rearranging gives the desired result, $$\color{#df0000}{\boxed{\int_0^\infty \frac{x \,dx}{e^x - 1} = \frac{\pi^2}{6}}} .$$
Best Answer
"Integrability" (more specifically, Lebesgue integrability) is a technical condition similar to absolute convergence of a sequence where a function is only considered integrable on $(a,b)$ if $$ \int_a^b|f(x)|dx < \infty.$$ By comparison with $1/x,$ $\sin(x)/x$ fails this criterion on $(-\infty, \infty)$ and is thus considered a non-integrable function on $(-\infty, \infty).$ This is likely the sense that was meant when you read that $\sin(x)/x$ is not integrable.
However, just like series can be convergent but not absolutely convergent, it turns out that since $\sin(x)/x$ oscillates as $|x|\rightarrow\infty$, the 'area under the curve' cancels out over the oscillations and is finite. The integral $$ \int_{-\infty}^\infty \frac{\sin(x)}{x}dx$$ can be defined as an improper Riemann integral and happens to equal $\pi.$ This is just a warning that 'not integrable' (in the Lebesgue sense) doesn't mean that the definite integral doesn't have a value.
All of this has little to do with your concern of finding an antiderivative for $\sin(x)/x$ and thus computing the integral over an interval using the fundamental theorem of calculus. As others have mentioned, it turns out that the antiderivative cannot be expressed as an elementary function, so it's no surprise you haven't been successful in applying various integration tricks.
Even though you cannot compute the integral in the usual way by taking an antiderivative, remember that, since $\sin(x)/x$ is a nice bounded, continuous function, the definite integral $\int_a^b\frac{\sin(x)}{x}dx$ exists a well-defined number (area under the curve) for any $a$ and $b$. Likewise, as others have mentioned, an antiderivative function $\mathrm{Si}(x) = \int_0^x\frac{\sin(t)}{t}dt$ exists and we have $$\int_a^b\frac{\sin(x)}{x}dx = \mathrm{Si}(b)-\mathrm{Si}(a).$$ It's just that $\mathrm{Si}(x)$ does not have a nice expression in terms of elementary functions like $\sin$, $\cos,$ $\ln$, $e^x,$ etc like you would usually compute with integration techniques.... it's its 'own thing' (called the sine-integral function).
Functions without an elementary antiderivative are usually called 'functions without an elementary antiderivative' and not called 'not integrable', although the latter would be perfectly reasonable terminology if it weren't already reserved for the technical condition I mentioned above.