I am currently juggling some integrals. In a physics textbook, Chaikin-Lubensky [1], Chapter 6, (6.1.26), I came upon an integral that goes \begin{equation} \int_0^{1} \textrm{d} y\, \frac{1 - J_0(y)}{y} - \int_{1}^{\infty} \textrm{d} y\, \frac{J_0(y)}{y} = -.116. \end{equation} They give the result only as a floating point value without naming sources. The value looks suspiciously like $\gamma - \ln(2)$ to me ($\gamma$ being the Euler-Mascheroni constant), which would solve a problem I have elsewhere. I am unfamiliar with the typical manipulations one uses on this kind of integrals and the various definitions of the Euler-Mascheroni constant. I fumbled around a bit with cosine integrals $\textrm{Ci}(y)$ but did not get far with it. So I am happy about suggestions.


5 Answers 5


A relatively elementary way is to start with known $$\gamma=\int_0^1\frac{1-\cos t}{t}\,dt-\int_1^\infty\frac{\cos t}{t}\,dt.$$

Put $t=ax$ for $a>0$ and do some rearrangements, to get $$\int_0^1\frac{1-\cos ax}{x}\,dx-\int_1^\infty\frac{\cos ax}{x}\,dx=\gamma+\log a.$$

Now the integral representation $J_0(y)=\frac2\pi\int_0^{\pi/2}\cos(y\cos x)\,dx$ yields $$\int_0^1\frac{1-J_0(y)}{y}\,dy-\int_1^\infty\frac{J_0(y)}{y}\,dy=\frac2\pi\int_0^{\pi/2}(\gamma+\log\cos x)\,dx$$ after interchanging integrations (which is not hard to justify).

The result now follows from $\int_0^{\pi/2}\log\cos x\,dx\color{gray}{=\int_0^{\pi/2}\log\sin x\,dx}=-(\pi/2)\log2$.

  • $\begingroup$ Nice and simple and (+1) $\endgroup$ Jun 10, 2022 at 11:52

For a wide class of function $f$, the following heuristic computation can be justified rigorously:1)

\begin{align*} &\int_{0}^{\infty} \frac{f(x) - f(0)\mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \left( f(x) - f(0)\mathbf{1}_{[0,1]}(x) \right) \left( \int_{0}^{\infty} e^{-xs} \, \mathrm{d}s \right) \, \mathrm{d}x \\ &= \int_{0}^{\infty} \left[ \int_{0}^{\infty} \left( f(x) - f(0)\mathbf{1}_{[0,1]}(x) \right) e^{-xs} \, \mathrm{d}x \right] \, \mathrm{d}s \tag{“Fubini”} \\ &= \int_{0}^{\infty} \left[ \mathcal{L}\{f\}(s) - f(0)\mathcal{L}\{\mathbf{1}_{[0,1]}\}(s) \right] \, \mathrm{d}s \\ &= \lim_{R \to \infty} \int_{0}^{R} \left[ \mathcal{L}\{f\}(s) - f(0)\mathcal{L}\{\mathbf{1}_{[0,1]}\}(s) \right] \, \mathrm{d}s \\ &= \lim_{R\to\infty} \left( \int_{0}^{R} \mathcal{L}\{f\}(s) \, \mathrm{d}s - f(0)(\gamma + \log R) \right) \end{align*}

where $\mathcal{L}\{f\}$ denotes the Laplace transform of $f$ and we utilized the well-known result that

$$ \int_{0}^{R} \mathcal{L}\{\mathbf{1}_{[0,1]}\}(s) \, \mathrm{d}s = \int_{0}^{R} \frac{1 - e^{-s}}{s} \, \mathrm{d}s = \gamma + \log R + o(1) $$

as $ R \to \infty $. (Hint: Perform integration by parts and use $\int_{0}^{\infty} e^{-s}\log s \, \mathrm{d}s = -\gamma$.) So, when $f = J_0$ is the Bessel function of the 1st kind and order $0$,

$$ J_0(0) = 1 \qquad \text{and}\qquad \mathcal{L}\{J_0\}(s) = \frac{1}{\sqrt{s^2 + 1}}, $$

and so,

\begin{align*} \int_{0}^{\infty} \frac{J_0(x) - \mathbf{1}_{(0,1)}(x)}{x} \, \mathrm{d}x &=\lim_{R\to\infty} \left[ \int_{0}^{R} \frac{\mathrm{d}s}{\sqrt{s^2+1}} - (\gamma + \log R) \right] \\ &=\lim_{R\to\infty} \left[ \log(R + \sqrt{R^2 + 1}) - (\gamma + \log R) \right] \\ &= \log 2 - \gamma. \end{align*}

Rearranging this yields

$$ \int_{0}^{\infty} \frac{\mathbf{1}_{(0,1)}(x) - J_0(x)}{x} \, \mathrm{d}x = \gamma - \log 2. $$

  • 1
    $\begingroup$ Really delightful! In the "well-known result" line, should you have $\int_0^R \mathcal{L}\{\mathbf{1}_{[0,1]}\}(s) ds$? $\endgroup$
    – user196574
    Jun 10, 2022 at 21:02
  • $\begingroup$ @user196574 Aww snap, my bad. Thank you for pointing out my mistake! $\endgroup$ Jun 10, 2022 at 21:12

Using the Meijer G-function $$\int \frac{J_0(y)}{y}\,dy=-\frac{1}{2} G_{1,3}^{2,0}\left(\frac{y^2}{4}| \begin{array}{c} 1 \\ 0,0,0 \end{array} \right)$$ $$\int_0^1 \frac{1-J_0(y)}{y}\,dy=\frac{1}{8} \, _2F_3\left(1,1;2,2,2;-\frac{1}{4}\right)$$ $$\int_1^\infty \frac{J_0(y)}{y}\,dy=\frac{1}{8} \, _2F_3\left(1,1;2,2,2;-\frac{1}{4}\right)-\gamma +\log (2)$$ Nice cancellation of terms !


Making it more general $$\int_0^a \frac{1-J_0(y)}{y}\,dy=\frac{1}{8} a^2 \, _2F_3\left(1,1;2,2,2;-\frac{a^2}{4}\right)$$ $$\int_a^\infty \frac{J_0(y)}{y}\,dy=\frac{1}{8} a^2 \, _2F_3\left(1,1;2,2,2;-\frac{a^2}{4}\right)-\log (a)-\gamma +\log (2)$$ $$\int_0^a \frac{1-J_0(y)}{y}\,dy-\int_a^\infty \frac{J_0(y)}{y}\,dy=\gamma+\log \left(\frac{a}{2}\right)$$

  • $\begingroup$ Thanks a lot, that looks quite elegant indeed. I have to work thorugh the definitions. $\endgroup$
    – Grob
    Jun 10, 2022 at 9:32
  • $\begingroup$ @Grob. Replace the denominator by $y^2$ and the result is $\left(1-\frac{1}{a}\right)$ $\endgroup$ Jun 10, 2022 at 9:41
  • $\begingroup$ Not sure why, but I derived the hypergeometric term. Just to show how it could be done in the general case $\endgroup$
    – Yuriy S
    Jun 10, 2022 at 11:01
  • $\begingroup$ @Grob. Replace the denominator by $\sqrt y $ and the result is $2 \sqrt{a}-\frac{\Gamma \left(\frac{1}{4}\right)}{\sqrt{2} \Gamma \left(\frac{3}{4}\right)}$ $\endgroup$ Jun 10, 2022 at 12:09
  • $\begingroup$ Excellent work! $\endgroup$
    – K.defaoite
    Jun 11, 2022 at 23:47

I'll try for a more direct proof.

$$ I=\int_0^{1} \textrm{d} y\, \frac{1 - J_0(y)}{y} - \int_{1}^{\infty} \textrm{d} y\, \frac{J_0(y)}{y} $$

We have:

$$J_0(y)=\frac{2}{\pi}\int_0^1 \frac{\cos (x y) dx}{\sqrt{1-x^2}}$$

Let's try (formal) series expansion of the cosine function, then we get:

$$J_0(y)=\frac{2}{\pi} \sum_{n=0}^\infty \frac{(-1)^n y^{2n}}{(2n)!} \int_0^1 \frac{x^{2n} dx}{\sqrt{1-x^2}}$$

Now we can (formally) subsitute this series into the original integral, I'll consider the first part for now:

$$I_1=\frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^{n-1} }{(2n)!} \int_0^1 \frac{x^{2n} dx}{\sqrt{1-x^2}} \int_0^1 y^{2n-1} dy$$

$$I_1=\frac{1}{\pi} \sum_{n=1}^\infty \frac{(-1)^{n-1} }{(2n)! n} \int_0^1 \frac{x^{2n} dx}{\sqrt{1-x^2}}$$

The last integral is just Beta function:

$$\int_0^1 \frac{x^{2n} dx}{\sqrt{1-x^2}}=\frac{1}{2} \int_0^1 \frac{u^{n-1/2} du}{\sqrt{1-u}}=\frac{1}{2} B\left(n+\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2} \frac{\Gamma(n+1/2)\Gamma(1/2)}{\Gamma(n+1)}=\\=\frac{\sqrt{\pi}}{2} \frac{\Gamma(n+1/2)}{n!}$$

This gives us:

$$I_1=\frac{1}{2\sqrt{\pi}} \sum_{n=1}^\infty \frac{(-1)^{n-1} \Gamma(n+1/2)}{(2n)! n! n}= \frac{1}{2\sqrt{\pi}} \sum_{n=0}^\infty \frac{(-1)^{n} \Gamma(n+3/2)}{(2n+2)! (n+1)! (n+1)}$$

This is clearly generalized hypergeometric series. Let's find the 0th order coefficient:


Now the ratio of coefficients:

$$\frac{c_{n+1}}{c_n}=\frac{(n+3/2)(n+1)(n+1)}{(2n+3)(2n+4)(n+2)(n+2)} \frac{-1}{n+1}=$$

$$=\frac{(n+1)(n+1)}{(n+2)(n+2)(n+2)} \frac{-1}{4(n+1)}$$

From this we conclude:

$$I_1=\frac18 {_2F_3} \left(1,1;2,2,2; -\frac14 \right)$$

I think the second integral can be done in a similar way, but there will be some limits required, which is where the constants come in.

I'll continue when I have the time.

  • $\begingroup$ Nice solution for sure $\endgroup$ Jun 10, 2022 at 11:51

Ramanujan's master theorem shows that $$\int_{0}^{\infty} J_{0}(t) t^{s-1} \, \mathrm dt= \frac{2^{s-1}\Gamma \left(\frac{s}{2} \right)}{\Gamma \left(1-\frac{s}{2} \right)}, \quad 0 < \Re(s) < 3/2. \tag{1}$$

The function $ J_{0}(t)-1$ has the same Mellin transform, but the region of convergence is $-2 < \Re(s) < 0$.

(See theorem 8.1 from the paper Ramanujan's Master Theorem by T. Amdeberhan et al.)

Therefore, we have$$ \begin{align} \int_{0}^{\infty} \frac{\mathbf{1}_{[0,1]}(t)-J_{0}(t)}{t} dt &=\lim_{s \to 0^{-}} \int_{0}^{\infty}\left( \mathbf{1}_{[0,1]}(t)-J_{0}(t) \right)t^{s-1} \, \mathrm dt \\&= \lim_{s \to 0^{-}} \left( \int_{0}^{\infty}\left(1-J_{0}(t) \right) t^{s-1} \, \mathrm dt - \int_{1}^{\infty} t^{s-1} \, \mathrm dt\right) \\ &= \lim_{s \to 0^{-}} \left( -\frac{2^{s-1}\Gamma \left(\frac{s}{2} \right)}{\Gamma \left(1-\frac{s}{2} \right)} + \frac{1}{s}\right) \\ &= \lim_{s \to 0^{-}} \left(\frac{-2^{s} \Gamma \left(1+\frac{s}{2}\right)+\Gamma \left(1- \frac{s}{2} \right)}{s \, \Gamma \left(1- \frac{s}{2} \right)}\right) \\ &\overset{(1)}{=} \lim_{s \to 0^{-}} \frac{- 2^{s} \ln(2) \Gamma\left(1+ \frac{s}{2} \right) - 2^{s-1} \Gamma'\left(1+ \frac{s}{2}\right) - \frac{1}{2} \Gamma'\left(1- \frac{s}{2} \right)}{\Gamma \left(1- \frac{s}{2} \right) - \frac{s}{2}\Gamma'\left(1- \frac{s}{2} \right)} \\ & = \frac{-\ln(2) - \frac{1}{2} \Gamma'(1)- \frac{1}{2} \Gamma'(1)}{1}\\ & = \gamma - \ln(2) . \end{align}$$

$(1)$ L'Hôpital's rule

The same approach shows that $$\int_{0}^{\infty} \frac{\frac{t}{2}\mathbf{1}_{[0,1]}(t)- J_{1}(t)}{t^{2}} \, \mathrm dt = \frac{2 \gamma - 2 \ln(2) -1}{4}. $$


$$ \begin{align} \int_{0}^{\infty} x^{s-1} \sum_{k=0}^{\infty} \frac{1}{\Gamma(k+2)^{2}} \left(- \frac{x^{2}}{4} \right)^{k} \, \mathrm dx &= 2^{s-1} \int_{0}^{\infty} u^{s/2-1} \sum_{k=0}^{\infty} \frac{1}{\Gamma(k+2)^{2}} \left(-u \right)^{k} \, \mathrm du \\ &= 2^{s-1} \, \frac{1}{\sin\left( \frac{\pi s}{2} \right) } \frac{1}{\Gamma \left(2-\frac{s}{2} \right)^{2}} \end{align}$$

Replacing $s$ with $s+2$, we have

$$ \begin{align} \int_{0}^{\infty} x^{s+2-1} \sum_{k=0}^{\infty} \frac{1}{\Gamma(k+2)^{2}} \left(- \frac{x^{2}}{4} \right)^{k} \, \mathrm dx &= - 4 \int_{0}^{\infty} x^{s-1} \sum_{k=0}^{\infty} \frac{1}{\Gamma(k+2)^{2}} \left(- \frac{x^{2}}{4} \right)^{k+1} \, \mathrm dx \\ &= \color{red}{- 4 \int_{0}^{\infty} x^{s-1} \sum_{k=1}^{\infty} \frac{1}{\Gamma(k+1)^{2}} \left(- \frac{x^{2}}{4} \right)^{k} \, \mathrm dx} \\ &= 2^{s+2-1} \, \frac{\pi}{\sin \left(\frac{\pi(s+2)}{2} \right)} \frac{1}{\Gamma\left(2- \frac{s+2}{2} \right)^{2} } \\ &= - 2^{s+1} \, \frac{\pi}{\sin \left(\frac{\pi s}{2} \right)} \frac{1}{\Gamma \left(1 - \frac{s}{2} \right)^{2}} \\ &= - 2^{s+1} \, \frac{\Gamma \left(\frac{s}{2} \right)}{\Gamma \left(1 -\frac{s}{2} \right)}. \end{align}$$


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