Integral of sin x/ x

The integral we’re setting out to solve here is the so-called Dirichlet integral:

    \[ \int_0^{\infty}\frac{\sin x }{x} \quad \text{(1)} \]

We can’t solve this integral by the usual means. There are multiple methods of solving this equation. 5 such methods are outlined at proofwiki. The method we’ll use here is Feynman’s technique. The proof is taken from:

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We need to find a way to get the x out of the denominator in the integrand. We do this as follows:

    \begin{align*} \int_0^{\infty}\frac{\sin x }{x} &= \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-0x}\,dx = I(0) \quad \text{(2)} \\ I(b) &= \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-bx}\,dx \quad \text{(3)} \\ \frac{d}{db}[I(b)] &= \frac{\partial}{\partial b}\left( \frac{\sin x }{x}\cdot e^{-bx} \right)\, dx \quad \text{(4)} \\ I^{\prime}(b) &= \frac{\partial}{\partial b}\displaystyle \left(\frac{\sin (x)\,e^{-bx}}{x}\right)\,dx \quad \text{(5)} \\ &= \displaystyle \left(-\frac{\cancel{x}\sin (x)\,e^{-bx}}{\cancel{x}}\right)\,dx \quad \text{(6)} \end{align*}

Now we’ve got the x out of the denominator of the integrand. Next, we need to integrate -\sin x \cdot e^{-bx}.

We use integration by parts twice in a row. Recall, in integration by parts:

    \[ \int fg^{\prime} = fg - \int f^{\prime}g \]

First Integration by Parts:

f=\sin(x)

f^{\prime}=\cos(x)

g^{\prime}=e^{-bx}

g=\displaystyle -\frac{e^{-bx}}{b}

So,

    \[\int_0^{\infty} e^{-bx}\,\sin(x)\,dx=\displaystyle -\frac{e^{-bx}\sin(x)}{b}-\int_0^{\infty} -\frac{e^{-bx}\cos(x)}{b}\]

Second Integration by Parts:

f=\cos(x)

f^{\prime}=-\sin(x)

g^{\prime}=\displaystyle -\frac{e^{-bx}}{b}

g=\displaystyle \frac{e^{-bx}}{b^2}

Therefore,

    \begin{align*} \int e^{-bx}\,\sin(x)\,dx &= -\frac{e^{-bx}\sin(x)}{b} - \frac{e^{bx}\cos(x)}{b^2} - \int_0^{\infty} \frac{e^{-bx}\sin(x)}{b^2}\,dx \\ \int \frac{e^{-bx}\sin(x)}{b^2}\,dx + \int e^{-bx}\,\sin(x)\,dx &= -\frac{e^{-bx}\sin(x)}{b} - \frac{e^{bx}\cos(x)}{b^2} \\ \frac{1}{b^2} \cdot \int e^{-bx}\sin(x)\dx + \frac{b^2}{b^2} \cdot \int_0^{\infty} e^{-bx}\,\sin(x)\,dx &= -\frac{e^{-bx}\sin(x)}{b} - \frac{e^{bx}\cos(x)}{b^2} \\ \frac{1 + b^2}{b^2} \cdot \left(\int e^{-bx}\sin(x)\,dx \right) &= -\frac{e^{-bx}\sin(x)}{b} - \frac{e^{bx}\cos(x)}{b^2} \\ \int e^{-bx}\sin(x)\,dx &= \frac{1 + b^2}{b^2} \cdot \left( -\frac{e^{-bx}\sin(x)}{b} - \frac{e^{bx}\cos(x)}{b^2} \right) \\ \int e^{-bx}\sin(x)\,dx &= -\frac{be^{-bx}\sin(x)+e^{bx}\cos(x)}{1 + b^2} \\ \int e^{-bx}\sin(x)\,dx &= -\frac{e^{-bx}(b\sin(x) + \cos(x))}{1 + b^2} \\ \int_0^{\infty} e^{-bx}\sin(x)\,dx &= \eval{[-\frac{e^{-bx}(b\sin(x) + \cos(x))}{1 + b^2}]}_0^{\infty} \\ &= -\frac{e^{-b(\infty)}(b\sin(x) + \cos(x))}{1 + b^2} - -\frac{e^{-b0}(b\sin(0) + \cos(0))}{1 + b^2} \\ &= -\frac{(0)(b\sin(x) + \cos(x))}{1+b^2} + \frac{(1)(0+1)}{1+b^2} \\ &= 0 + \frac{1}{1+b^2} \\ &= \frac{1}{1+b^2} \end{align*}

In this derivation, to make things simpler, I’ve left off the addition of constants for indefinite integrals and I did the integral of \displaystyle +\,\int \sin x \cdot e^{-bx} rather than \displaystyle -\,\int \sin x \cdot e^{-bx} so I have to put the (-) sign back in. Our final result, then, is:

    \[ -\int_0^{\prime} \sin x \cdot e^{-bx} = -\frac{1}{1+b^2} \]

    \[ I^{\prime}(b) = -\frac{1}{1+b^2} \quad \text{(7)} \]

We integrate both sides:

    \[ \int I^{\prime}(b) = I(b) = -\int \frac{1}{1+b^2} = -\arctan(b) + C \quad \text{(8)} \]

(To see how to get this result, click here.)

Eq. (3) tells us that:

    \[ I(b) = \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-bx}\,dx \quad \text{(9)} \]

Thus:

    \[ I(b) = -\arctan(b) + C = \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-bx}\,dx \quad \text{(10)} \]

Take the limit as b\rightarrow \infty:

    \begin{align*} -\arctan(\infty) + C &= \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-(\infty)x}\,dx \quad \text{(11)} \\ -\frac{\pi}{2} + C &= 0 \quad \text{(12)} \\ C &= \frac{\pi}{2} \quad \text{(13)} \\ I(b) &= -\arctan(b) + \frac{\pi}{2} \quad \text{(14)} \\ I(0) &= -\arctan(0) + \frac{\pi}{2} \\ &= 0 + \frac{\pi}{2} \\ &= \frac{\pi}{2} \quad \quad \quad \quad \quad \quad \quad \text{(15)} \end{align*}

But:

    \begin{align*} I(b) &= \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-bx}\,dx \quad \text{(16)} \\ I(0) &= \int_0^{\infty}\frac{\sin x }{x}\cdot e^{-0x}\,dx \\ &= \int_0^{\infty}\frac{\sin x }{x}\cdot (1)\,dx \\ &= \int_0^{\infty}\frac{\sin x }{x}\,dx \\ &= \frac{\pi}{2} \quad \quad \quad \quad \quad \quad \quad \quad \text{(17)} \end{align*}

That means that:

    \[ \int_0^{\infty}\frac{\sin x }{x}\,dx = \frac{\pi}{2} \quad \text{(18)} \]