# Is $\int_0^1{\frac{\sin x}{x}dx}$ convergent?

I have this integral:

$$\int_0^1{\frac{\sin x}{x}dx}$$ And I should prove that it is convergent. I have understand that if the resulting area is finite, then this integral is convergent, right?

So, I do this:

$$\int_0^1{\frac{\sin x}{x}dx} = \int_0^1{\sin x\cdot \frac{1}{x}dx} =\\= \begin{bmatrix} -\cos x\cdot \ln x\end{bmatrix}_0^1 = (-\cos1\cdot \ln1)-(\cos0\cdot \ln0) = 0-(-\infty) = \infty$$

But, I get an infinit value, which I think is wrong. What am I doing wrong?

Ok, so apparently I messed some things up. If I do this again and use the integration by parts method I get this:

$$\int_0^1{\frac{\sin x}{x}dx} = \int_0^1{\sin x\cdot \frac{1}{x}dx} = \sin x \ln x- \int_0^1{(\cos x\ln x) dx} = \sin x \ln x -\begin{bmatrix} \sin\frac{1}{x} \end{bmatrix}_0^1 \Rightarrow \lim_{t \rightarrow 0} {\int_t^1\frac{\sin x}{x} dx} = \lim_{t \rightarrow 0}{(\sin x \ln x-\frac{sin1}{1}-\frac{\sin t}{t})}$$

Is that more correct? And how do I proceed?

• You are making believe that $\int f(x)g(x)=\int f(x)\int g(x)$. Commented Feb 24, 2014 at 10:01
• Defining the integrand to have value $1$ at $x=0$ makes it continuous on $[0,1]$. Commented Feb 24, 2014 at 10:02
• Did you just say that $$\int f(x)g(x)dx = \int f(x)dx = \int g(x)dx?$$
– 5xum
Commented Feb 24, 2014 at 10:02
• Hint: In the region of interest, $\sin(x)$ is positive and $\sin(x)\leq x$. Commented Jan 2, 2017 at 14:48

If we define the function f(x) by

f(x) = sin(x) / x if x is not 0 and 1 if x is 1

then f is a continuous function. Therefore f can be integral

I think that we can't express explicit integral value as we known such algebraic number or transcendental number or etc.

You must know that:

• The integral of a product is not the product of integrals;

• If you modify an integrable function in a finite number of points, the function remains integrable and the integral remains the same;

• The function $\frac{\sin x}{x}$ is not defined in $x=0$ but has a finite limit for $x\to 0$. Hence it can be etended to a continuous function defined on the whole interval $[0,1]$; The integral on $(0,1]$ is equal to the integral of such function on $[0,1]$, and this integral is finite.

I would also be useful to know that:

• the integral of the function $\frac{\sin x }{x}$ cannot be written explicitly in terms of elementary functions (i.e. algebraic, logaritmic and trigonometric functions)

The integral $$I = \int_{0}^{1}\frac{\sin x}{x}\,dx$$ is a proper Riemann integral in contrast to the improper Riemann integral like $$J = \int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}} = \frac{\pi}{2}$$ because in case of integral $I$ the integrand as well as the interval of integration is bounded (for $J$ the interval of integration is bounded, but the integrand is not). The issue of convergence / divergence of integral appears only in case of improper Riemann integrals and thus is not applicable to the integral $I$ in question here.

Unfortunately the function $f(x) = (\sin x)/x$ does not have an elementary anti-derivative (it is somewhat difficult to prove this fact) and hence it is useless to expect a closed form representation of $I$ using elementary functions. The integral $I$ can be evaluated numerically to any desired level of accuracy by using suitable approximation methods.

Rewrite

$$\lim_{\epsilon \rightarrow0}\bigg(\int_0^\epsilon f(x)+\int_\epsilon^1f(x)\bigg)$$

Now split the limit and use:

$$\lim_{\epsilon \rightarrow0} \int_0^\epsilon f(x)\le \lim_{\epsilon \rightarrow0} \epsilon \sup_{x\in[0,\epsilon]}|f(x)|\rightarrow0$$

Now since

$$\int_\epsilon^1f(x)$$

is bounded ($f(x)$ is well defined on $[\epsilon,1]$ ) you are done...

Since $\sin x = x+O(x^3)$, $\sin x/x = 1+O(x^2)$, so the integral certainly exists.