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Recently, I found these two interesting integrals in Handbook of special functions page 141. $$\mathcal{I}=\int_{0}^{\infty }\frac{\sin(ax)}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$$ $$\mathcal{J}=\int_{0}^{\infty }\frac{\cos(ax)}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$$ In this book, it gives the answer below $$\mathcal{I}=\frac{\pi }{2}\left (\ln\frac{z}{2a}-\gamma \right )\left [ I_0\left ( az \right )- \mathbf{L}_0\left ( az \right )\right ]+\frac{1}{4\pi }G_{24}^{32}\left ( \frac{a^{2}z^{2}}{4}\middle|\begin{matrix} \dfrac{1}{2},\dfrac{1}{2} \\ 0,0,\dfrac{1}{2},\dfrac{1}{2} \end{matrix} \right )~~~,~~~\left (a,\Re z>0 \right )$$ $$\mathcal{J}=\left ( \ln\frac{z}{2a}-\gamma \right )K_0\left ( az \right )~~~,~~~\left ( a,\Re z>0 \right )$$ where $I_0(\cdot)$ is modified bessel function of the first kind, $\mathbf{L}_0(\cdot)$ is modified struve function, $G_{pq}^{mn}(\cdot)$ is meijer-G function and $K_0(\cdot)$ is bessel function of rhe second kind.

So, I tried to figure out how to get the answer.

My attempt: Let $x=z\tan t$, we have \begin{align*} \mathcal{I}&=2\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\ln\left ( z\sec t \right )\sec t\, \mathrm{d}t\\ &=2\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\ln\left ( \sec t \right )\sec t\, \mathrm{d}t+2\ln z\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\sec t\, \mathrm{d}t \end{align*} Hence, define $$\mathcal{I}\left ( m \right )=\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\sec^mt\, \mathrm{d}t$$ then using the taylor series of $\sin x$ we get \begin{align*} \mathcal{I}\left ( m \right )&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k+1}}{\left ( 2k+1 \right )!}\int_{0}^{\frac{\pi }{2}}\tan^{2k+1}t\sec^mt\, \mathrm{d}t \\ &=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k+1}}{\left ( 2k+1 \right )!}\int_{0}^{\frac{\pi }{2}}\sin^{2k+1}t\cos^{-2k-m-1}t\, \mathrm{d}t \end{align*} By using the same way we get $$\begin{align*} \mathcal{J}\left ( m \right )&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k}}{\left ( 2k \right )!}\int_{0}^{\frac{\pi }{2}}\tan^{2k}t\sec^mt\, \mathrm{d}t \\ &=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k}}{\left ( 2k \right )!}\int_{0}^{\frac{\pi }{2}}\sin^{2k}t\cos^{-2k-m}t\, \mathrm{d}t \end{align*}$$ But how to evaluate the last integral, it seems can't be expressed by Beta function.

If I'm doing the wrong way, is there another way to solve the problem.

Any help will be appreciated!

  • 0
    perhaps $\sin^2(t) \to \rho$ and $\cos^2(t) \to (1-\rho)$ will help – 2017-01-17
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    the first integral with the Meijer-G isn't really appealing to me – 2017-01-17
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    For the second integral, have you tried something along this lines? http://math.stackexchange.com/questions/1954866/is-there-a-way-to-compute-int-0-infty-frac-cos-qt-j-1-qr1q2-ma – 2017-01-17
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    Furthermore the prefactor in front of $K_0(z) $ looks suspicious like its small $z$ expansion – 2017-01-17
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    And I was thinking $K_0$ is the elliptic integral of the first kind. – 2017-01-17
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    Do some scaling first to bring the integral into standard form. Then note that the integrand is even. Define a complex valued function $f(z)=(1+z^2)^{\nu}e^{iqz}$ and integrate it over a semicircle in the complex plane with a slit $(i,i\infty)$. Doing everything correctly you will find a famous integral representation for $ K_{\nu}(q)$ times an elementary function (there are no more contributions due to analyticality). Now take $\partial_{\nu}K_{\nu}(q)$ according to DMLF 10.38 and take limit $\nu\rightarrow 0$ in the end. http://dlmf.nist.gov/10.38 – 2017-01-17
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    @tired Thanks! I'll try! – 2017-01-17

1 Answers 1

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For the second integral

Note that

$$ K_\nu(az)=\frac{\Gamma(\nu+1/2)(2z/a)^\nu}{\sqrt{\pi}}\int_0^\infty\frac{\cos at }{(t^2+z^2)^{\nu+1/2}} dt$$

By differentiation with respect to $\nu$

\begin{align} \frac{\partial K_\nu(az)}{ \partial \nu} &=(\Gamma'(\nu+1/2)+\log(2z/a) )K_\nu(az)\\&-\frac{\Gamma(\nu+1/2)(2z/a)^\nu}{\sqrt{\pi}}\int_0^\infty\frac{\cos a t }{(t^2+z^2)^{\nu+1/2}} \log(x^2+z^2)dt \end{align}

Note that in

$$\left|\frac{\partial K_\nu(z)}{ \partial \nu} \right|_{\nu=0} = 0$$

This implies

$$\int_0^\infty\frac{\cos a t }{\sqrt{t^2+z^2)}} \log(x^2+z^2)\,dt = (\Gamma'(1/2)+\log(2z/a) )K_0(az) $$

Note that

$$\Gamma'(1/2)+\log(2z/a) = -\gamma -2\log(2)+\log(2)+\log(z/a) =\log(z/2a) -\gamma$$

Hence

$$\mathcal{J}=\left ( \log (z/2a)-\gamma \right )K_0\left ( az \right )$$

Addendum

$$K_\nu (z) = \int^\infty_0 e^{-z\cosh t} \cosh(\nu t)\,dt$$

This implies

$$\frac{\partial K_\nu(z)}{ \partial \nu} = \int^\infty_0t e^{-z\cosh t} \sinh(\nu t)\,dt$$

Hence we have

$$\left|\frac{\partial K_\nu(z)}{ \partial \nu} \right|_{\nu=0} = 0$$

  • 0
    uhhh, nice i didn't know this particular integral rep.... an additional proof of your last statement would be awesome! – 2017-01-17
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    Thx! But for the first one, it seems more difficult! Does it have a similar method like this? – 2017-01-17
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    @tired, see the addendum. – 2017-01-17
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    @Renascence_5 , I think we can apply it to the first integral but this will be tedious. – 2017-01-17