# Solving $\int_0^{\infty} \ln^m(x)\sin\left(x^n\right)\:dx$

Spurred on by this question, I decided to investigate a more generalised form:

$$$$I_{m,n} = \int_0^{\infty} \ln^m(x)\sin\left(x^n\right)\:dx$$$$

Where $$n,m \in \mathbb{N}$$

I have formed a solution in terms of the Gamma Function but I'm unsure whether it can be expressed in terms of other Non-Elementary and/or Elementary Functions. Also very interested to see other approaches (Real + Complex Analysis).

To solve, we first observe that:

$$$$I_{n,k} = \lim_{\phi\rightarrow 0^+} \frac{d^m}{d\phi^m}\int_0^\infty x^\phi \sin\left(x^n\right)\:dx$$$$

Here let:

$$$$J_{n}(\phi) = \int_0^\infty x^\phi \sin\left(x^n\right)\:dx$$$$

We observe that we first must solve $$J_{n,k}(\phi)$$. To achieve we employ Feynman's Trick coupled with Laplace Transforms. This is allowable as the integrand conforms with both Fubini's Theorem and the Dominated Convergence Theorem. Here we introduce:

$$$$H_{n}(t,\phi) = \int_0^\infty x^\phi \sin\left(tx^n\right)\:dx$$$$

Where

$$$$J_{n}(\phi) = \lim_{t\rightarrow 1^+} H_{n}(t,\phi)$$$$

We now take the Laplace Transform of $$H_{n}(t,\phi)$$ with respect to $$t$$:

\begin{align} \mathscr{L}_t\left[H_{n}(t,\phi)\right] = \int_0^\infty x^\phi \mathscr{L}_t\left[\sin\left(tx^n\right)\right]\:dx = \int_0^\infty x^\phi \frac{x^n}{s^2 + x^{2n}} \:dx = \int_0^\infty \frac{x^{\phi + n}}{s^2 + x^{2n}} \:dx \end{align}

Thankfully (and as I address here) this integral can be evaluated easily: \begin{align} \mathscr{L}_t\left[H_{n}(t,\phi)\right] = \int_0^\infty \frac{x^{\phi + n}}{s^2 + x^{2n}} \:dx = \frac{1}{2n} \cdot \left(s^2\right)^{ \frac{\phi + n + 1}{2n} - 1}\cdot B\left(1 - \frac{\phi + n + 1}{2n}, \frac{\phi + n + 1}{2n} \right) \end{align} Using the relationship between the Beta Function and the Gamma Function:

$$$$\mathscr{L}_t\left[H_{n}(t,\phi)\right] = \frac{1}{2n} s^{ \frac{\phi + n + 1}{n} - 2}\Gamma\left(1 - \frac{\phi + n + 1}{2n} \right)\Gamma\left(\frac{\phi + n + 1}{2n} \right)$$$$

We now resolve $$H_{n}(t, \phi)$$ by taking the Inverse Laplace Transform:

\begin{align} H_{n}(t,\phi)&=\mathscr{L}_s^{-1}\left[ \frac{1}{2n} s^{ \frac{\phi + n + 1}{n} - 2}\Gamma\left(1 - \frac{\phi + n + 1}{2n} \right)\Gamma\left(\frac{\phi + n + 1}{2n} \right)\right]\\ & = \frac{1}{2n} \cdot \frac{1}{\Gamma\left(2 - \frac{\phi + n + 1}{n}\right)t^{-\left(\frac{\phi + n + 1}{n} - 2 + 1\right)} } \cdot \Gamma\left(1 - \frac{\phi + n + 1}{2n} \right)\Gamma\left(\frac{\phi + n + 1}{2n} \right) \end{align}

We can now solve $$J_n(\phi)$$:

$$$$J_{n}(\phi) = \lim_{t\rightarrow 1^+} H_{n}(t,\phi) = \frac{\Gamma\left(1 - \frac{\phi + n + 1}{2n} \right)\Gamma\left(\frac{\phi + n + 1}{2n} \right) }{2n\:\Gamma\left(2 - \frac{\phi + n + 1}{n}\right) }$$$$

And finally we have

$$$$I_{m,n} = \int_0^{\infty} \ln^m(x)\sin\left(x^n\right)\:dx = \lim_{\phi\rightarrow 0^+} \frac{d^m}{d\phi^m} \left[\frac{\Gamma\left(1 - \frac{\phi + n + 1}{2n} \right)\Gamma\left(\frac{\phi + n + 1}{2n} \right) }{2n\:\Gamma\left(2 - \frac{\phi + n + 1}{n}\right) } \right]$$$$

For example, using the example as linked above we have $$m = 2$$, $$n = 2$$:

$$$$I_{2,2} = \int_0^{\infty} \ln^2(x)\sin\left(x^2\right)\:dx = \lim_{\phi\rightarrow 0^+} \frac{d^2}{d\phi^2} \left[\frac{\Gamma\left(1 - \frac{\phi + 2 + 1}{2\cdot 2} \right)\Gamma\left(\frac{\phi + 2 + 1}{2\cdot2} \right) }{2n\:\Gamma\left(2 - \frac{\phi + 2 + 1}{2}\right) } \right]$$$$

I was too lazy to do it by hand, but evaluated through WolframAlpha we observe that:

$$$$I_{2,2} = \int_0^{\infty} \ln^2(x)\sin\left(x^2\right)\:dx = \frac{1}{32}\sqrt{\frac{\pi}{2}}(2\gamma-\pi+4\ln2)^2$$$$

As required

• You are leaps and bounds ahead of me in terms of integration skill at this point. Really great work David. – clathratus Jan 31 at 16:04
• @clathratus - Thank you for your kind words. It's just a matter of seeing where I can apply the general case I found a few weeks back. – user150203 Feb 1 at 9:50

A method relying on the Mellin transform of the sine:

For $$s>1$$ and $$z \in \mathbb{C}$$ with $$-s < \operatorname{Re} (z) < s$$ we have $$f_s (z) \equiv \int \limits_0^\infty x^{z-1} \sin(x^s) \, \mathrm{d} x = \frac{1}{s} \int \limits_0^\infty t^{\frac{z}{s}-1} \sin(t) \, \mathrm{d} t = \frac{1}{s} \mathcal{M}(\sin) \left(\frac{z}{s}\right) = \frac{1}{s} \sin \left(\frac{\pi z}{2s}\right) \operatorname{\Gamma} \left(\frac{z}{s}\right)$$ with the limit $$f_s(0) = \frac{\pi}{2s}$$. For $$m \in \mathbb{N_0}$$ and $$s>1$$ this implies \begin{align} I_{m,s} &\equiv \int \limits_0^\infty \ln^m(x) \sin(x^s) \, \mathrm{d} x = f_s^{(m)}(1) = \frac{1}{s} \frac{\mathrm{d}^m}{\mathrm{d} z^m} \left[\sin \left(\frac{\pi z}{2s}\right) \operatorname{\Gamma} \left(\frac{z}{s}\right)\right] \Bigg\vert_{z=1} \\ &= \frac{1}{s^{m+1}} \frac{\mathrm{d}^m}{\mathrm{d} x^m} \left[\sin \left(\frac{\pi }{2}x\right) \operatorname{\Gamma} \left(x\right)\right] \Bigg\vert_{x= 1/s} = \frac{1}{s^{m+1}} \sum \limits_{k=0}^m {m \choose k} \left(\frac{\pi}{2}\right)^k \sin^{(k)} \left(\frac{\pi}{2s}\right) \operatorname{\Gamma}^{(m-k)} \left(\frac{1}{s}\right) \, , \end{align} where the last step follows from the general Leibniz rule. The derivatives of the gamma function can be expressed in terms of polygamma functions using Faà di Bruno's formula, but otherwise that is probably as elementary as it gets.

In the special case $$s=2$$ we can use $$\sin^{(k)} \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} (-1)^{\lfloor k/2 \rfloor}$$ for $$k \in \mathbb{N}_0$$ and the values (obtained from the Legendre duplication formula) \begin{align} \operatorname{\Gamma} \left(\frac{1}{2}\right) &= \sqrt{\pi} \, , \\ \operatorname{\psi}^{(0)} \left(\frac{1}{2}\right) &= - \gamma - 2 \ln(2) \, , \\ \operatorname{\psi}^{(n)} \left(\frac{1}{2}\right) &= (-1)^{n-1} n! (2^{n+1}-1) \zeta(n+1) \, , \, n \in \mathbb{N} \, , \end{align} to simplify the final result. The complexity of Faà di Bruno's formula prevents us from finding a reasonably nice general expression for $$(I_{m,2})_{m \in \mathbb{N}_0}$$ , but at least we know that these integrals can be written in terms of $$\pi$$, $$\ln(2)$$, $$\gamma$$ and zeta values.

• A Fantastic Solution. Thanks for posting (+1) – user150203 Feb 1 at 9:46

Using your own parameterized integral $$J_n(\phi)$$ we may use Ramanujan's Master Theorem.

Ramanujan's Master Theorem $$($$RMT$$)$$

Let $$f(x)$$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\varphi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\varphi(-s)$$

For the purpose of exploiting this theorem we expand the sine function as its corresponding MacLaurin Series and enforce the substitution $$x^n\mapsto x$$ firstly and $$x^2\mapsto x$$ afterwards $$($$Note: the distinction between $$x^n\mapsto x$$ and $$x^2\mapsto x$$ instead of directly $$x^{2n}\mapsto x$$ is only chosen for simplicity$$)$$. This leads to

\begin{align*} J_n(\phi)=\int_0^\infty x^{\phi}\sin(x^n)\mathrm dx &= \int_0^\infty x^{\phi}\left[\sum_{k=0}^\infty(-1)^k\frac{(x^n)^{2k+1}}{(2k+1)!}\right]\mathrm dx\\ &=\int_0^\infty x^{\phi/n}\left[\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}\right]\frac{\mathrm dx}{nx^{1-1/n}}\\ &=\frac1n\int_0^\infty x^{(\phi+1)/n}\left[\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k+1)!}\right]\mathrm dx\\ &=\frac1n\int_0^\infty x^{(\phi+1)/2n}\left[\sum_{k=0}^\infty(-1)^k\frac{x^k}{(2k+1)!}\right]\frac{\mathrm dx}{2x^{1/2}}\\ &=\frac1{2n}\int_0^\infty x^{(\phi+1)/2n-1/2}\sum_{k=0}^\infty\frac{\Gamma(k+1)/\Gamma(2k+2)}{k!}(-x)^k\mathrm dx \end{align*}

The new gained structure is cleary recognizable as possible application of the RMT. Therefore set $$s=\frac{\phi+1}{2n}+\frac12$$ and moreover $$\varphi(k)=\frac{\Gamma(k+1)}{\Gamma(2k+2)}$$. Thus, we get

\begin{align*} J_n(\phi)&=\frac1{2n}\int_0^\infty x^{(\phi+1)/2n-1/2}\sum_{k=0}^\infty\frac{\Gamma(k+1)/\Gamma(2k+2)}{k!}(-x)^k\mathrm dx\\ &=\frac1{2n}\Gamma\left(\frac{\phi+1}{2n}+\frac12\right)\frac{\Gamma\left(1-\left(\frac{\phi+1}{2n}+\frac12\right)\right)}{\Gamma\left(2-2\left(\frac{\phi+1}{2n}+\frac12\right)\right)}\\ &=\frac1{2n}\frac{\Gamma\left(\frac12+\frac{\phi+1}{2n}\right)\Gamma\left(\frac12-\frac{\phi+1}{2n}\right)}{\Gamma\left(1-\frac{\phi+1}n\right)}\\ &=\frac1{2n}\frac1{\Gamma\left(1-\frac{\phi+1}n\right)}\frac{\pi}{\sin\left(\frac{\phi+1}{2n}\pi+\frac\pi2\right)}\\ &=\frac1{n}\frac1{\Gamma\left(1-\frac{\phi+1}n\right)}\frac{\pi}{2\cos\left(\frac{\phi+1}{2n}\pi\right)}\frac{\sin\left(\frac{\phi+1}{2n}\pi\right)}{\sin\left(\frac{\phi+1}{2n}\pi\right)}\\ &=\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right) \end{align*}

$$\therefore~J_n(\phi)~=~\int_0^\infty x^{\phi}\sin(x^n)\mathrm dx~=~\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right)$$

From hereon we can deduce the same formulae ComplexYetTrivial's answer contains, and which he already did quite well by invoking Leibniz rule and Faà di Bruno's formula. The crucial point of this post is to present another possible derivation in order to obtain the Mellin Transform of the sine function. In my opionion using the RMT explains the close connection to the Gamma Function quite well. Of course, one has to be careful with the choice of $$\phi$$ and $$n$$ hence an occuring negative integer value within on of the Gamma Functions in the nominator would cause an indefinite expression.

$$\therefore~I_{m,n}~=~\int_0^\infty \ln^m(x)\sin(x^n)\mathrm dx~=~\lim_{\phi\to0}\frac{\mathrm d^m}{\mathrm d\phi^m}\left[\frac1n\Gamma\left(\frac{\phi+1}n\right)\sin\left(\frac{\phi+1}{2n}\pi\right)\right]$$

• For your final expression, is it meant to be the mth derivative? – user150203 Feb 1 at 0:17
• @DavidG Yes, of course. I forgot to write down the $m$. I will fix this right now. – mrtaurho Feb 1 at 6:14
• Just wanted to check (worried I may have missed something). Btw, love your solution (+1) – user150203 Feb 1 at 9:46