# A tough integral : $\int_{0}^{\infty }\frac{\sin x \text{ or} \cos x}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$

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!

• perhaps $\sin^2(t) \to \rho$ and $\cos^2(t) \to (1-\rho)$ will help – Francesco Alem. Jan 17 '17 at 8:53
• the first integral with the Meijer-G isn't really appealing to me – tired Jan 17 '17 at 8:55
• For the second integral, have you tried something along this lines? math.stackexchange.com/questions/1954866/… – tired Jan 17 '17 at 9:45
• Furthermore the prefactor in front of $K_0(z)$ looks suspicious like its small $z$ expansion – tired Jan 17 '17 at 9:58
• And I was thinking $K_0$ is the elliptic integral of the first kind. – Zaid Alyafeai Jan 17 '17 at 10:11

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

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

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

This implies

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

Hence we have

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

• uhhh, nice i didn't know this particular integral rep.... an additional proof of your last statement would be awesome! – tired Jan 17 '17 at 10:43
• Thx! But for the first one, it seems more difficult! Does it have a similar method like this? – Renascence_5. Jan 17 '17 at 10:59
• @tired, see the addendum. – Zaid Alyafeai Jan 17 '17 at 11:28
• @Renascence_5 , I think we can apply it to the first integral but this will be tedious. – Zaid Alyafeai Jan 17 '17 at 11:29