While trying to solve a challenging system of dual integral equations resulting from a mixed boundary value problem arising in a fluid mechanical problem, the four following non-trivial convergent improper integrals emerge: \begin{align} I_1 (r,t) &= \int_0^\infty \lambda^{\frac{1}{2}} e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda \, , \\ I_2 (r,t) &= \int_0^\infty \lambda^{\frac{1}{2}} e^{-\lambda} J_0 (\lambda r) J_{\frac{3}{2}} (\lambda t) \, \mathrm{d} \lambda\, , \\ I_3 (r,t) &= \int_0^\infty \left( \lambda^{\frac{1}{2}}-\lambda^{-\frac{1}{2}} \right) e^{-\lambda} J_1 (\lambda r) J_{\frac{3}{2}} (\lambda t) \, \mathrm{d}\lambda \, , \\ I_4 (r,t) &= \int_0^\infty \left( \lambda^{\frac{1}{2}}+\lambda^{-\frac{1}{2}} \right) e^{-\lambda} J_0 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda \, , \end{align} wherein $t$ and $r$ are positive real numbers. It can be checked that, thanks to the exponential function, these integrals are convergent.

If the integrands do not contain the exponential function, then the evaluation of these integrals is easy and straightforward.

Is there probably a way to evaluate these integrals analytically even as infinite convergent series functions? Any help or hint is highly appreciated!

  • 1
    $\begingroup$ I wouldn't hope for any closed forms, but expressions involving hypergeometric series are certainly achievable. Try starting with $$J_\alpha(z)=\frac{(z/2)^\alpha}{\Gamma(\alpha+1)}\,_0F_1\left(;\alpha+1;-\tfrac{z^2}{4}\right),$$ and $$\left(\sum_{n\ge0}a_nx^n\right)\left(\sum_{n\ge0}b_nx^n\right)=\sum_{n\ge0}x^n\sum_{u+v=n}a_ub_v,$$ so you can expand each integrand into uniformly convergent power series, and interchange the sum and integral. $\endgroup$
    – clathratus
    Commented Nov 15, 2019 at 7:54
  • 1
    $\begingroup$ Are you missing a $t$ in the exponentials ? $\endgroup$ Commented Nov 15, 2019 at 8:02
  • 1
    $\begingroup$ For $t=r$, there are solutions. Otherwise ??? $\endgroup$ Commented Nov 15, 2019 at 9:36
  • 1
    $\begingroup$ Yes, I know ! This is why I wrote "???". Cheers :-) $\endgroup$ Commented Nov 15, 2019 at 10:04
  • 1
    $\begingroup$ @Volterra In general, $$_pF_q(a_1,a_2,...,a_p;b_1,b_2,...,b_q;z)=\sum_{n\ge0}\frac{(a_1)_n(a_2)_n\cdots (a_p)_n}{(b_1)_n(b_2)_n\cdots (b_q)_n}\frac{z^n}{n!},$$ where $(x)_n=\Gamma(x+n)/\Gamma(x)$. The function $_pF_q$ is the generalized hypergeometric function. $\endgroup$
    – clathratus
    Commented Nov 15, 2019 at 15:09

2 Answers 2


I believe these integrals have a simple analytical form. I will demonstrate for $I_1$ and I hope you can see how to do the others similarly.

I write $I_1$ out as originally stated:

$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t)$$

Note that

$$J_{1/2}(\lambda t) = \sqrt{\frac{2}{\pi \lambda t}} \sin{(\lambda t)}$$ $$J_1(\lambda r) = \frac1{i \pi} \int_0^{\pi} d\theta \, \cos{\theta} \, e^{i \lambda r \cos{\theta}} $$

Plugging back into the integral definition of $I_1$ and changing the order of integration, we get

$$I_1 = \frac1{i \pi} \sqrt{\frac{2}{\pi t}} \int_0^{\pi} d\theta \, \cos{\theta} \, \int_0^{\infty} d\lambda \, e^{-\lambda} \, \sin{(\lambda t)} \, e^{i \lambda r \cos{\theta}} $$

Rewriting the sine in exponential form, the integral over $\lambda$ is simple, and we are left with the integral over $\theta$:

$$I_1 = -\frac1{2 \pi} \sqrt{\frac{2}{\pi t}} \int_0^{\pi} d\theta \, \cos{\theta} \, \left [\frac1{1-i t - i r \cos{\theta}} - \frac1{1+i t - i r \cos{\theta}} \right ] $$

Now let's consider the integral

$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} $$

where $a$ and $b$ may be complex; in our case $a=1\pm i t$ and $b=-i r$. While there are at least a couple of ways of evaluating this integral, I will demonstrate how it is done using contour integration.

Consider the contour integral

$$-i \oint_C \frac{dz}{z} \, \frac{z^2+1}{b z^2+2 a z+b} $$

where $C$ is the following contour:

enter image description here

The semicircle has unit radius. Note that, because the real integral is only over a half-cycle rather than a full cycle, the contour $C$ includes a traversal along the real axis. Nevertheless, because of the pole at the origin, there needs to be a small detour of radius $\epsilon$ around the origin as shown.

The contour integral is then equal to

$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - i \, PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} - i (i \epsilon) \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac1{\epsilon \, e^{i \phi}} \frac{\epsilon^2 e^{i 2 \phi}+1}{b \epsilon^2 e^{i 2 \phi}+ 2 a \epsilon \, e^{i \phi} + b} $$

The first integral is what we seek (for now). The third integral is, in the limit as $\epsilon \to 0$, $-\pi/b$. The second integral, the principal value integral, may be evaluated as follows:

$$\begin{align} PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} &= \frac1{b} \, PV \int_{-1}^1 \frac{dx}{x}\, \left (1 - \frac{2 a x}{b x^2+2 a x+b} \right ) \\ &= \frac1{b} \, PV \int_{-1}^1 \frac{dx}{x} - \frac{2 a}{b} \int_{-1}^1 \frac{dx}{b x^2+2 a x+b}\end{align}$$

Note that the first principal value integral on the RHS vanishes by symmetry. The second integral on the right needs not be expressed using principal value notation because the pole at the origin is removed. Accordingly,

$$\begin{align} PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} &= - \frac{2 a}{b} \int_{-1}^1 \frac{dx}{b x^2+2 a x+b} \\ &= -\frac{2 a}{b^2} \int_{-1}^1 \frac{dx}{\left ( x+\frac{a}{b} \right )^2 + 1-\frac{a^2}{b^2}}\\ &= -\frac{2 a}{b^2} \frac1{\sqrt{1-\frac{a^2}{b^2}}} \left [ \arctan{\left ( \frac{x+\frac{a}{b}}{\sqrt{1-\frac{a^2}{b^2}}} \right )} \right ]_{-1}^1 \\ &= \frac{\pi}{b} \frac{a}{\sqrt{b^2-a^2}} \end{align}$$

For convenience later on, we may write

$$ PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} = -i \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} $$

This way, we may write that the contour integral is equal to

$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} - \frac{\pi}{b} $$

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand of the contour integral inside the contour $C$. In this case, the only pole inside the contour is at $z_+ = -\frac{a}{b} + \sqrt{\frac{a^2}{b^2}-1}$. Computing the residue at this pole, the resulting equation for the integral we seek for now is

$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} - \frac{\pi}{b} = -\frac{2 \pi}{b} \frac{a}{\sqrt{a^2-b^2}}$$


$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} = -\frac{\pi}{b} \left ( \frac{a}{\sqrt{a^2-b^2}} - 1 \right ) $$

Now we may use this result to determine $I_1$. Again, subbing $a=1 \pm i t$ and $b=-i r$, we get that

$$\begin{align} I_1 &= -\frac1{2 \pi} \sqrt{\frac{2}{\pi t}} \frac{\pi}{i r} \left ( \frac{1-i t}{\sqrt{(1-i t)^2+r^2}} - \frac{1+i t}{\sqrt{(1+i t)^2+r^2}} \right ) \\ &= \sqrt{\frac{2}{\pi t}} \frac1{r} \operatorname{Im}{\left (\frac{1+i t}{\sqrt{(1+i t)^2+r^2}} \right )}\end{align}$$

And with that, we are technically finished. But as someone who likes explicit results, I will take this a bit further and express the result as follows:

$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t) = \\ \frac1{\sqrt{\pi t r^2}} \frac{t \sqrt{\sqrt{(1+r^2-t^2)^2+4 t^2}+(1+r^2-t^2)} - \sqrt{\sqrt{(1+r^2-t^2)^2+4 t^2}-(1+r^2-t^2)}}{\sqrt{(1+r^2-t^2)^2+4 t^2}}$$

I have verified this in Mathematica numerically.

  • $\begingroup$ Thanks Ron a lot for your formidable answer. In the 4th and 5th equations in your answer $\sqrt{\lambda}$ should not appear in front of the integral. $\endgroup$
    – keynes
    Commented Nov 16, 2019 at 10:40
  • $\begingroup$ @Volterra: Thanks for catching the typo. All fixed now. $\endgroup$
    – Ron Gordon
    Commented Nov 16, 2019 at 10:47

Laplace transform can be used to evaluate these integrals. We have the Laplace transforms \begin{align} &\mathcal{L}\left[J_0(x)\right](p)=\frac{1}{\sqrt{1+p^2}};\quad \mathcal{L}\left[J_1(x)\right](p)=1-\frac{p}{\sqrt{1+p^2}}\\ &\mathcal{L}\left[J_2(x)\right](p)=\frac{1}{\sqrt{1+p^2}\left( p+\sqrt{1+p^2} \right)^2};\quad\mathcal{L}\left[J_3(x)\right](p)=\frac{1}{\sqrt{1+p^2}\left( p+\sqrt{1+p^2} \right)^3} \end{align} the recurrence relations \begin{equation} J_{3/2}(z)=\frac{1}{z}J_{1/2}(z)-J_{-1/2}(z);\quad \frac{1}{\lambda}J_1\left( \lambda r \right)=\frac{r}{2}\left[J_0\left( \lambda r \right)+J_2\left( \lambda r \right)\right] \end{equation} and the properties \begin{equation} J_{1/2}(\lambda t) = \sqrt{\frac{2}{\pi \lambda t}} \sin{(\lambda t)};\quad J_{-1/2}(\lambda t) = \sqrt{\frac{2}{\pi \lambda t}} \cos{(\lambda t)} \end{equation}

  • Then, for $I_1$, \begin{align} I_1 (r,t) &= \int_0^\infty \lambda^{\frac{1}{2}} e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda \\ &=\sqrt{\frac{2}{\pi t}}\int_0^\infty e^{-\lambda} J_1 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda \\ &=\sqrt{\frac{2}{\pi t}}\Im\left[\frac{1}{r}\int_0^\infty e^{-\frac{x}{r}\left( 1-it \right)} J_1 (x) \, \mathrm{d}x \right]\\ &=\sqrt{\frac{2}{\pi t}}\frac{1}{r}\Im\left[\frac{1+it}{\sqrt{r^2+(1+it)^2}}\right] \end{align} which is the @RonGordon result.

  • The integral $I_4$ can be obtained in the same way, by using an integral representation for $\lambda^{-1}\sin\lambda t$, we find \begin{align} I_4 (r,t) &= \int_0^\infty \left( \lambda^{\frac{1}{2}}+\lambda^{-\frac{1}{2}} \right) e^{-\lambda} J_0 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda \\ &=\sqrt{\frac{2}{\pi t}}\int_0^\infty \left( 1+\frac{1}{\lambda} \right)e^{-\lambda} J_0 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda \\ &=\sqrt{\frac{2}{\pi t}}\left[\int_0^\infty e^{-\lambda} J_0 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda+ \int_0^\infty \frac{e^{-\lambda}}{\lambda} J_0 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda \right]\\ &=\sqrt{\frac{2}{\pi t}}\left[\int_0^\infty e^{-\lambda} J_0 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda+\int_0^t\,d\tau\int_0^\infty e^{-\lambda} J_0 (\lambda r)\cos{(\lambda \tau)} \, \mathrm{d}\lambda\right]\\ &=\sqrt{\frac{2}{\pi t}}\left[\Im\left[\frac{1}{\sqrt{r^2+(1-it)^2}}\right]+\int_0^t\,d\tau\Re\left[\frac{1}{\sqrt{r^2+(1-i\tau)^2}}\right]\right]\\ &=\sqrt{\frac{2}{\pi t}}\left[\Im\left[\frac{1}{\sqrt{r^2+(1-it)^2}}\right]+\Re\left[\arcsin\left( \frac{t+i}{r} \right)\right]\right] \end{align} The real part of $\arcsin$ can be explicitly expressed (see here, for example).

For the other two integrals, we use the recurrence relation on $J_{3/2}$.

  • $I_2$ can be expressed as \begin{align} I_2 (r,t) &= \int_0^\infty \lambda^{\frac{1}{2}} e^{-\lambda} J_0 (\lambda r) J_{\frac{3}{2}} (\lambda t) \, \mathrm{d} \lambda\\ &=\frac{1}{t}\int_0^\infty \lambda^{-\frac{1}{2}} e^{-\lambda} J_0 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d} \lambda- \int_0^\infty \lambda^{\frac{1}{2}} e^{-\lambda} J_0 (\lambda r) J_{-\frac{1}{2}} (\lambda t) \, \mathrm{d} \lambda \end{align} Noticing that the first integral was calculated while evaluating $I_4$ and that
    \begin{equation} J_{-1/2}(\lambda t) = \sqrt{\frac{2}{\pi \lambda t}} \cos{(\lambda t)} \end{equation} we have directly \begin{equation} I_2=\sqrt{\frac{2}{\pi t^3}}\Re\left[\arcsin\left( \frac{t+i}{r} \right)\right]-\sqrt{\frac{2}{\pi t}}\Re\left[\frac{1}{\sqrt{r^2+(1-it)^2}}\right] \end{equation}

  • For $I_3$, \begin{align} I_3 (r,t) &= \int_0^\infty \left( \lambda^{\frac{1}{2}}-\lambda^{-\frac{1}{2}} \right)e^{-\lambda} J_1 (\lambda r) J_{\frac{3}{2}} (\lambda t) \, \mathrm{d}\lambda \\ &=\frac{1}{t}\int_0^\infty \lambda^{-\frac{1}{2}}e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda+ \int_0^\infty \lambda^{-\frac{1}{2}}e^{-\lambda} J_1 (\lambda r) J_{-\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda\\ &-\int_0^\infty \lambda^{\frac{1}{2}}e^{-\lambda} J_1 (\lambda r) J_{-\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda -\frac{1}{t}\int_0^\infty \lambda^{-\frac{3}{2}}e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda\\ &=I_{3,1}+I_{3,2}-I_{3,3}-I_{3,4} \end{align} The first integral, \begin{align} I_{3,1}&=\frac{1}{t}\int_0^\infty \lambda^{-\frac{1}{2}}e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda\\ &=\sqrt{\frac{2}{\pi t^3}}\int_0^\infty \frac{e^{-\lambda}}{\lambda} J_1 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda \\ &=\sqrt{\frac{2}{\pi t^3}}\int_0^t\,d\tau\int_0^\infty e^{-\lambda} J_1 (\lambda r)\cos{(\lambda t)} \, \mathrm{d}\lambda\\ &=\sqrt{\frac{2}{\pi t^3}}\frac{1}{r}\int_0^t\,d\tau\Re\left[1-\frac{1-i\tau}{\sqrt{r^2+(1-i\tau)^2}} \right]\\ &=\sqrt{\frac{2}{\pi r^2t^3}}\left[t+\Im\left[\sqrt{r^2+(1-i\tau)^2}\right]\right] \end{align} The second one, \begin{align} I_{3,2}&=\int_0^\infty \lambda^{-\frac{1}{2}}e^{-\lambda} J_1 (\lambda r) J_{-\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda\\ &=\sqrt{\frac{2}{\pi t}}\int_0^\infty \frac{e^{-\lambda}}{\lambda} J_1 (\lambda r)\cos{(\lambda t)} \, \mathrm{d}\lambda \\ &=\sqrt{\frac{r^2}{2\pi t}}\left[\int_0^\infty e^{-\lambda} J_0 (\lambda r)\cos{(\lambda t)} \, \mathrm{d}\lambda +\int_0^\infty e^{-\lambda} J_2 (\lambda r)\cos{(\lambda t)} \, \mathrm{d}\lambda\right]\\ &=\sqrt{\frac{r^2}{2\pi t}}\Re\left[ \frac{1}{\sqrt{r^2+(1-it)^2}}+ \frac{r^2}{\sqrt{r^2+(1-it)^2}\left( 1-it+\sqrt{r^2+(1-it)^2} \right)^2} \right] \end{align}

The third one can be deduced from the calculation of $I_1$, by taking the real part: \begin{equation} I_{3,3}=\sqrt{\frac{2}{\pi t r^2}}\Re\left[1-\frac{1-it}{\sqrt{r^2+(1-it)^2}}\right] \end{equation} The fourth one is slightly more complicated: \begin{align} I_{3,4}&=\frac{1}{t}\int_0^\infty \lambda^{-\frac{3}{2}}e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda\\ &=\sqrt{\frac{2}{\pi t^3}}\int_0^\infty \frac{e^{-\lambda}}{\lambda^2} J_1 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda\\ %&=\sqrt{\frac{2}{\pi t^3}}\int_0^t\,d\tau\int_0^\infty \frac{e^{-\lambda}}{\lambda} J_1 (\lambda r)\cos{(\lambda \tau)} \, \mathrm{d}\lambda\\ &=\sqrt{\frac{r^2}{2\pi t^3}}\left[ \int_0^\infty \frac{e^{-\lambda}}{\lambda}J_0 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda +\int_0^\infty \frac{e^{-\lambda}}{\lambda}J_2 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda\right]\\ &=\sqrt{\frac{r^2}{2\pi t^3}}\left[\int_0^t\,d\tau \int_0^\infty e^{-\lambda}J_0 (\lambda r)\cos{(\lambda \tau)} \, \mathrm{d}\lambda +\frac{r}{4}\int_0^\infty e^{-\lambda}J_1 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda +\frac{r}{4}\int_0^\infty e^{-\lambda}J_3 (\lambda r)\sin{(\lambda t)} \, \mathrm{d}\lambda\right]\\ &=\sqrt{\frac{r^2}{2\pi t^3}}\left[ \int_0^t\,d\tau \Re\left[\frac{1}{\sqrt{r^2+(1-i\tau)^2}}\right] +\frac{1}{4}\Im\left[1-\frac{1-it}{\sqrt{r^2+(1-it)^2}}\right] +\frac{r}{4}\Im\left[\frac{r^3}{\sqrt{r^2+(1-it)^2}\left( 1-it+\sqrt{r^2+(1-it)^2} \right)^3}\right] \right] \end{align} it comes \begin{align} I_{3,4}&=\sqrt{\frac{r^2}{2\pi t^3}}\left[ \Re\left[\arcsin\left( \frac{t+i}{r} \right)\right] -\frac{1}{4}\Im\left[\frac{1-it}{\sqrt{r^2+(1-it)^2}}\right] +\frac{r^4}{4}\Im\left[\frac{1}{\sqrt{r^2+(1-it)^2}\left( 1-it+\sqrt{r^2+(1-it)^2} \right)^3}\right] \right] \end{align} These expression seem to be numerically correct.

  • $\begingroup$ Thanks Paul, please try to find out what could be wrong with $I_3$. Greetings $\endgroup$
    – keynes
    Commented Nov 20, 2019 at 7:06
  • $\begingroup$ $I_2$ does not seem to be correct, i am afraid... $\endgroup$
    – keynes
    Commented Nov 20, 2019 at 7:40
  • 1
    $\begingroup$ I checked the expressions and I confirm them. I have $\arcsin((t+i)/r)=0.3307271996 + 2.844174493\,i$ and $(r^2 + (1 - it)^2)^{-1/2}=0.8897757790 + 0.3034321479\,i$. $\endgroup$
    – Paul Enta
    Commented Nov 20, 2019 at 9:05
  • 1
    $\begingroup$ All the expressions for $I_3$ are corrected now (there were mistakes) and they were checked numerically. I also tried to improve the readability... $\endgroup$
    – Paul Enta
    Commented Nov 20, 2019 at 15:27
  • 1
    $\begingroup$ Fantastic, thanks. Yes, the 3 last terms of $I_3$ where wrong... i will also double check these expression now myself! $\endgroup$
    – keynes
    Commented Nov 20, 2019 at 15:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .