How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$

Question

How to prove the integral $$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$

I try to use integration by parts $$\begin{align} &\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\ =&\int_0^\infty \int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\\ =&x\left.\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right|_0^\infty+\int_0^\infty \cos x \int_x^\infty\frac{\sin t}{t}dx+\int_0^\infty \sin x \int_x^\infty\frac{\cos t}{t}dx \end{align}$$ But the last two integrals are divergent.

Answer 1

Well, you can show that:

$$\int\limits_0^\infty\text{Si}\left(x\right)\text{Ci}\left(x\right)\space\text{d}x=\frac{1}{4\pi}\int\limits_0^\infty\left(\frac{\ln\left(1+\text{s}^2\right)}{\text{s}}\right)^2\space\text{ds}=\ln\left(2\right)\tag1$$

Using evaluating integrals over the positive real axis property of the Laplace transform.

Answer 2

The Laplace transform is useful for providing integral representations in terms of non-oscillating functions: $$\text{si}(x)=\int_{x}^{+\infty}\frac{\sin t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(\cos(x)+s\sin(x)\right)\,ds $$ $$\text{Ci}(x)=\int_{x}^{+\infty}\frac{\cos t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(s\cos(x)-\sin(x)\right)\,ds $$ This allows to apply Fubini's theorem: $$\begin{eqnarray*} \int_{0}^{+\infty}\text{si}(x)\text{Ci}(x)\,dx &=& \iiint_{(0,+\infty)^3}\frac{e^{-(a+b)x}}{(1+a^2)(1+b^2)}(\cos x+a\sin x)(b\cos x-\sin x)\,dx\,da\,db\\&=&\iint_{(0,+\infty)^2}\frac{1}{(1+a^2)(1+b^2)}\left(1-\frac{a}{a+b}+\frac{ab-3}{4+(a+b)^2}\right)\,da\,db\end{eqnarray*} $$ and the last double integral is pretty simple to be computed: $$ \iint_{(0,+\infty)^2}\frac{da\,db}{(1+a^2)(1+b^2)}=\left(\frac{\pi}{2}\right)^2 $$ $$ \iint_{(0,+\infty)^2}\frac{a}{(1+a^2)(1+b^2)(a+b)}\,da\,db=\iint_{(0,+\infty)^2}\frac{b}{(1+a^2)(1+b^2)(a+b)}\,da\,db =\frac{\pi^2}{8}$$ and the difficult part $$ \iint_{(0,+\infty)^2}\frac{ab-3}{(1+a^2)(1+b^2)(4+(a+b)^2)}\,da\,db=\int_{0}^{+\infty}\frac{-\pi(a^2+1)+2(a^2-1)\arctan\frac{a}{2}+2a\log(a^2+4)}{4(a^2+1)^2}\,da$$ can be broken by integration by parts.

Answer 3

Integration by parts shows that $$ \int \operatorname{Ci}(x) \, \mathrm dx = x \operatorname{Ci}(x) - \int \cos (x) \, \mathrm dx = x \operatorname{Ci}(x) - \sin(x) +C_{1} .$$

Then again using integration by parts, we have

$$ \begin{align} \int \operatorname{si}(x) \operatorname{Ci}(x) \, \mathrm dx &= \int \underbrace{\left(\operatorname{Si}(x)- \frac{\pi}{2} \right) }_{u} \underbrace{\operatorname{Ci}(x) \, \mathrm dx}_{dv} \\ &= \left(\operatorname{Si} (x)- \frac{\pi}{2} \right) \left( x\operatorname{Ci}(x) - \sin (x)\right) - \int \frac{\sin (x)}{x}\left(x\operatorname{Ci}(x) - \sin (x) \right) \, \mathrm dx, \end{align}$$

where $$ \begin{align} \int \underbrace{\operatorname{Ci}(x))}_{u} \underbrace{\sin(x)\, \mathrm dx}_{dv} &= - \cos(x) \operatorname{Ci}(x) + \int \frac{\cos^{2}(x)}{x} \, \mathrm dx \\ &= - \cos(x) \operatorname{Ci}(x) + \frac{1}{2} \int \left(\frac{1}{x} + \frac{\cos (2x)}{x} \right) \, \mathrm dx \\ &=- \cos(x) \operatorname{Ci}(x) + \frac{\ln (x)}{2} + \frac{\operatorname{Ci}(2x)}{2} + C_{2} \end{align}$$

and $$ \int \frac{\sin^{2}(x)}{x} \, \mathrm dx = \int \frac{1}{2} \left(\frac{1}{x}- \frac{\cos(2x)}{x} \right) \, \mathrm dx = \frac{\ln (x)}{2} - \frac{\operatorname{Ci}(2x)}{2} + C_{3}.$$

Therefore,

$$ \small \int \operatorname{si}(x) \operatorname{Ci}(x) \, \mathrm dx = \left(\operatorname{Si} (x)- \frac{\pi}{2} \right) \left( x\operatorname{Ci}(x) - \sin (x)\right)+ \cos(x) \operatorname{Ci}(x) - \operatorname{Ci}(2x) + C_{4} \tag{1}. $$

We can can use the asymptotic form $$\operatorname{Ci}(x) \sim \frac{\sin x}{x} $$ for large $x$ to show that the limit of the right side of $(1)$ as $x \to + \infty$ is $C_{4}$, and then use use the series expansion $$\operatorname{Ci}(x) = \gamma + \ln(x) + \mathcal{O}(x^{2})$$ to show that limit of the right side of $(1)$ as $x \to 0^{+}$ is $-\ln(2) + C_{4}$.

This leads to the result.

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Since $\int_{0}^{\infty} \operatorname{Ci}(x) \, \mathrm dx = 0$, it turns out that $$ \int_{0}^{\infty} \operatorname{si}(x) \operatorname{Ci}(x) \, \mathrm dx = \int_{0}^{\infty} \operatorname{Si}(x) \operatorname{Ci}(x) \, \mathrm dx. $$

Answer 4

Let $E_{1}(z)$ be the exponential integral function defined on the right half of the complex plane by the integral $$E_{1}(z)=\int_{1}^{\infty} \frac{e^{-zt}}{t} \, \mathrm dt, \quad \left( \Re(z) \ge 0, \ z \ne 0 \right). $$

Near the origin, $E_{1}(z)$ behaves like $- \ln(z)$.

And when $|z|$ is large in magnitude, $E_{1}(z)$ behaves like $\frac{e^{-z}}{z}$.

Therefore, since $|e^{-z}| \le 1$ when $\Re(z) \ge 0$, we can integrate $\left( E_{1}(z)\right)^{2}$ around a closed quarter circle in the first quadrant of the complex plane (indented at the origin) and conclude that $$\int_{0}^{\infty} \left( E_{1}(x)\right)^{2} \, \mathrm dx - \int_{0}^{\infty} \left(E_{1}(it)\right)^{2} \, \mathrm \, i \, dt =0. $$

But notice that for $t>0$, $$E_{1}(it) = \int_{1}^{\infty} \frac{\cos (tu)}{u} \, \mathrm du - i \int_{1}^{\infty} \frac{\sin(tu)}{u} \, \mathrm du = - \operatorname{Ci}(t) + i \operatorname{si}(t). $$

Therefore, $$\int_{0}^{\infty}\left(E_{1}(x)\right)^{2} \, \mathrm dx - i\int_{0}^{\infty} \left(\left( \operatorname{Ci}(t)\right)^{2}-2 i \operatorname{Ci}(t) \operatorname{si}(t) - \left(\operatorname{si}(t)\right)^{2}\right) \, \mathrm dt =0. $$

And equating the real parts on both sides of the above equation, we get $$ \begin{align} \int_{0}^{\infty} \operatorname{Ci}(t) \operatorname{si}(t) \, \mathrm dt &= \frac{1}{2} \int_{0}^{\infty} \left(E_{1}(x)\right)^{2} \, \mathrm dx \\ &= \int _{0}^{\infty} e^{-x} E_{1}(x) \, \mathrm dx \tag{1} \\ &= \int_{0}^{\infty} e^{-x} \int_{1}^{\infty} \frac{e^{-xv}}{v} \, \mathrm dv \, \mathrm dx \\ &= \int_{1}^{\infty} \frac{1}{v} \int_{0}^{\infty} e^{-(1+v)x} \, \mathrm dx \, \mathrm dv \\ &= \int_{1}^{\infty} \frac{\mathrm dv}{v(1+v)} \\ &= \ln 2. \end{align}$$

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$(1)$ Integrate by parts.