Integral $ \int_0^{\infty} \ln{x} \sin{ px}\ e^{-qx} dx$

Question

I am trying to figure out a closed form solutions of the following integral:

$$ \operatorname{I}\left(p,q\right) = \int_{0}^{\infty}\ln\left(x\right)\sin\left(px\right){\rm e}^{-qx}\,{\rm d}x\,, \quad p,q>0 $$

It is taken from "Table of Integrals, Series, and Products" $7_{\rm th}$ ed. by Gradshteyn and Ryzhik, where they state that on pg $599$, \begin{align} & \int_{0}^{\infty}{\rm e}^{-qx}\sin\left(px\right) \ln\left(x\right){\rm d}x \\[2mm] = & \ \frac{1}{p^{2} + q^{2}} \left[q\arctan\left(\frac{p}{q}\right) -pC -\frac{p}{c}\,\ln\left(p^{2} - q^{2}\right)\right] \end{align} without telling us about the constant $C$.

• The integral satisfies $$ \frac{\partial^{2}\operatorname{I}}{\partial p^{2}} = -p^{2}\operatorname{I}\quad\mbox{and}\quad \frac{\partial\operatorname{I}} {\partial q} = -q\operatorname{I} $$ but, without any inital conditions, these equations were not helpful.

Is there a way to verify this result and figure out the constant $C$ ?.

Answer 1

Use the integral $$\int_0^{\infty}e^{-qx}\ln{x}\left(\sin px-\frac{pq}{p^2+q^2}\right)dx=I+\frac{p(\gamma+\ln q)}{p^2+q^2}$$ Note that $\gamma$ is the Euler-Mascheroni constant. Integrate by parts with $u=\ln x$ and $dv$ equal to the other part and evaluate from there. $$\frac{1}{q^2+p^2}\int_0^\infty\frac{q\sin px+p\cos px-p}{x}e^{-qx}dx$$ $$=\frac{1}{q^2+p^2}\int_0^\infty\frac{q\sin px+p\cos px-p}{x}\int_q^\infty xe^{-xy}dy\space dx$$ $$=\frac{1}{q^2+p^2}\int_q^\infty\int_0^\infty ( q\sin px+p\cos px-p)e^{-xy}dx\space dy$$$$=\left(-\frac{p}{2}\ln (p^2+q^2)+p\ln(q)+q\arctan(\frac{p}{q})\right)\frac{1}{p^2+q^2}$$ Solving for I, we get $$I=\int_0^\infty e^{-qx}\ln{x}\sin px\space dx=\frac{1}{p^2+q^2}\left(-\frac{p}{2}\ln (p^2+q^2)-p\gamma+q\arctan(\frac{p}{q})\right)$$

Answer 2

It can also be obtained from the integral formula $$\int_{0}^{\infty} x^{s-1} \sin(px) e^{-qx} \, \mathrm dx = \frac{\Gamma(s)}{(p^{2}+q^{2})^{s/2}} \sin \left(s \arctan \frac{p}{q} \right), \quad (p, q>0, \ s>-1) $$ by differentiating under the integral sign.

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$ \begin{align} &\int_{0}^{\infty} \ln(x) \sin(p x) e^{-qx} \, \mathrm dx \\ &= \lim_{s \to 1} \frac{\partial }{\partial s}\int_{0}^{\infty}x^{s-1} \sin(px) e^{-qx} \, \mathrm dx\\ &= \lim_{s \to 1} \frac{\partial }{\partial s}\frac{\Gamma(s)}{(p^{2}+q^{2})^{s/2}} \, \sin \left(s \arctan \frac{p}{q} \right) \\ &= \small\lim_{s \to 1}\left(\frac{\Gamma'(s) \sin \left(s \arctan \frac{p}{q} \right)}{(p^{2}+q^{2})^{s/2}}- \frac{\Gamma(s) \ln(p^{2}+q^{2}) \sin \left(s \arctan \frac{p}{q} \right)}{2 (p^{2}+q^{2})^{s/2}}+ \frac{\Gamma(s) \arctan \left(\frac{p}{q} \right) \cos \left(s \arctan \frac{p}{q} \right)}{(p^{2}+q^{2})^{s/2}} \right) \\ &\overset{\diamondsuit}{=} \left( \frac{- \gamma \, \frac{p}{(p^{2}+q^{2})^{1/2}}}{(p^{2}+q^{2})^{1/2}}- \frac{\ln(p^{2}+q^{2}) \frac{p}{(p^{2}+q^{2})^{1/2}}}{2 (p^{2}+q^{2})^{1/2}} + \frac{\arctan \left(\frac{p}{q} \right)\frac{q}{(p^{2}+q^{2})^{1/2}}}{(p^{2}+q^{2})^{1/2}} \right) \\ &= \frac{1}{p^{2}+q^{2}} \left(- p \gamma -\frac{p}{2} \ln(p^{2}\color{red}{+}q^{2}) + q \arctan \left(\frac{p}{q} \right) \right) \end{align}$

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$\diamondsuit$ Derivative of Gamma Function at 1

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Similarly,

$$ \begin{align} \int_{0}^{\infty} \ln(x) \cos(p x) e^{-qx} \, \mathrm dx &= \lim_{s \to 1} \frac{\partial }{\partial s}\frac{\Gamma(s)}{(p^{2}+q^{2})^{s/2}} \, \cos \left(s \arctan \frac{p}{q} \right) \\ &= \frac{1}{p^{2}+q^{2}} \left(- q \gamma -\frac{q}{2} \ln(p^{2} + q^{2}) - p \arctan \left(\frac{p}{q} \right) \right). \end{align}$$

Answer 3

Integrate both \begin{align} I(p,q)=&\int_0^{\infty} e^{-qx} \ln{x} \sin px\ dx\\ J(p,q)=&\int_0^{\infty} e^{-qx} \ln{x} \cos px\ dx\\ \end{align} by parts, respectively, to establish

\begin{align} I(p,q)=&-\frac1{2p}\ln(p^2+q^2)-\frac{\lambda}p-\frac qp J(p,q)\\ J(p,q)=&-\frac1{p}\tan^{-1}\frac qp +\frac qp I(p,q)\\ \end{align} where the two simpler integrals \begin{align} &\int_0^{\infty} {\frac{e^{-qx}\sin p x}x}dx =\tan^{-1}\frac qp\\ & \int_0^{\infty} {\frac{e^{-qx} (1-\cos p x)}x}dx =\frac12\ln(1+\frac{p^2}{q^2}) \end{align} as well as $ \int_0^{\infty} e^{-x}\ln x\ dx =-\gamma$ are assumed known. Then, solve the system of the two equations above to obtain \begin{align} I(p,q)=&-\frac1{p^2+q^2}\left(\frac p2\ln(p^2+q^2)+p{\lambda}-q \tan^{-1}\frac qp \right)\\ J(p,q)=& -\frac1{p^2+q^2}\left(\frac q2\ln(p^2+q^2)+q{\lambda}+p \tan^{-1}\frac qp \right)\\ \end{align}