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My intention is to express the LHS in the RHS -form

$$\int \frac1{\sqrt{x}}\ln(x)\sin(x) dx=x \sqrt{x}\,\left[F(x)\ln(x)+G(x)\right] + C$$

where $x>0$. WA shows that the solution requires something-called gamma-fuctions and generalized hypergeometric functions, here. $F(x)$ and $G(x)$ are undefined. When I look at the WA -solution, I am bit flabbergasted what to believe because WA has more terms than the supposed. Could someone outline this problem? Is the problem just chain-rule or is there some easy-way to get into the right hand side?

I have started from series expansion but I am not sure wheher the right way. I have tried to research why gamma-function appears in this integral but cannot yet understand it, wikipedia here. I am not sure from WA -notation whether the gamma function at hand is in even defined by reals or complexes (I think in complexes now). In reals, it is apparently just a generalization of factorial. And here is the hypergeometric

$$_{2}F_{1}(a, b; c; z)=\sum_{n=0}^{\infty}\frac{(a)_{n} (b)_{n}}{(c)_{n}}\frac{z^{n}}{n!},$$

fully defined here.

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What kind of class is this homework for? I suspect there may be a mistake in the question, because I'm pretty sure this integral can't be expressed as an elementary function - you need special functions (Maple uses LommelS1 and its derivatives with respect to the first and second parameters). – Robert Israel Feb 5 '12 at 8:22
@RobertIsrael: It is an engineering course, page 538 here on the M-word-documented course book (sorry not in English). This question is jammed in the part serie-expanion and partial-fraction. It is very open-ended and misleading question with free functions $F(x)$ and $G(x)$ -- or it is possible that the $F(x)$ is defined so that: $\begin{cases}F'(x)=\frac{1}{\sqrt{2}}e^{\frac{-x^{2}}{2}}, x\in\mathbb R \\ F(0)=\frac{1}{2} \end{cases}$. Sorry this question is not one of the clearest and one can understand this in multiple ways. – hhh Feb 5 '12 at 8:38
up vote 2 down vote accepted

The idea of this homework is to identify some functions $F$ and $G$ such that the RHS is a solution. Note that the derivative of the RHS is $$ \tfrac32\sqrt{x}(F(x)\log(x)+G(x))+x\sqrt{x}(x^{-1}F(x)+F'(x)\log(x)+G'(x)), $$ Introduce the function $$ S(x)=\frac{\sin(x)}x=\sum\limits_{n\geqslant0}s_nx^{2n},\qquad s_n=\frac{(-1)^n}{(2n+1)!}. $$ Then the derivative of the RHS minus the LHS is $$ \sqrt{x}\left(\tfrac32G(x)+F(x)+xG'(x)\right)+\sqrt{x}\log(x)\left(\tfrac32F(x)+xF'(x)-S(x)\right),$$ hence the RHS is a solution as soon as $$ \tfrac32G(x)+F(x)+xG'(x)=0,\qquad \tfrac32F(x)+xF'(x)=S(x).$$ For the next step, it is probably best to use series expansions. Assume that $$ F(x)=\sum\limits_{n\geqslant0}a_nx^{2n},\quad G(x)=\sum\limits_{n\geqslant0}b_nx^{2n}. $$ The two first order differential equations above yield, for every $n\geqslant0$, $$ \tfrac32b_n+a_n+2nb_n=0,\qquad\tfrac32a_n+2na_n=s_n, $$ that is, $$ a_n=\frac{s_n}{2n+\frac32},\qquad b_n=\frac{-s_n}{(2n+\frac32)^2}. $$ This determines entirely some functions $F$ and $G$, written as series of infinite radius of convergence, such that the derivative of the RHS is the function in the integral. As a consequence, this determines completely the desired primitives on the domain $x\gt0$.

One can write the functions $F$ and $G$, hence the primitives themselves, in several equivalent ways. The one I prefer is the one above. But if one is fond of generalized hypergeometric functions, one can express $F$ and $G$ with these. For example, $F(x)=u\left(-\frac{x^2}4\right)$, with $$ u(z)=\sum\limits_{n\geqslant0}u_nz^n,\qquad u_n=\frac{4^n}{(2n+1)!(2n+\frac32)}. $$ To write $u_n$ as a ratio of Pochhammer symbols, the procedure is standard. Start from $$ \frac{u_{n+1}}{u_n}=\frac{4(2n+\frac32)}{(2n+3)(2n+2)(2n+\frac52)}=\frac{n+\frac34}{(n+\frac32)(n+\frac54)}\frac1{n+1}. $$ Since $u_0=\frac23$, $(a)_0=1$ and $(a)_{n+1}=(a)_n(a+n)$ for every $a$ and every $n\geqslant0$, this yields $$ u_n=\frac{\frac23(\frac34)_n}{(\frac32)_n(\frac54)_n}\frac1{n!}. $$ Finally, $$ F(x)=\frac23\,{}_1\!F_2\left(\frac34;\frac32,\frac54;-\frac{x^2}4\right). $$ Likewise, one can express $G$ in terms of a hypergeometric function of type ${}_2\!F_3$, an exercise we will leave to the reader.

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My guess is that a series expansion is wanted.
Start from the rhs. If $J(x) = x^{3/2} (F(x) \ln(x) + G(x)) + C$, then $J'(x) = \left(\frac{3}{2} x^{1/2} F(x) + x^{3/2} F'(x) \right) \ln(x) + x^{1/2} F(x) + \frac{3}{2} x^{1/2} G(x) + x^{3/2} G'(x)$. If $F(x) = f_0 + f_1 x + f_2 x^2 + \ldots$ and $G(x) = g_0 + g_1 x + g_2 x^2 + \ldots$, this is $$ \left(\frac{3}{2} f_0 x^{1/2} + \frac{5}{2} f_1 x^{3/2} + \frac{7}{2} f_2 x^{5/2} + \ldots\right) \ln(x) + \left(\frac{3}{2} g_0 + f_0\right) x^{1/2} + \left(\frac{5}{2} g_1 + f_1\right) x^{3/2} + \left(\frac{7}{2} g_2 + f_2\right) x^{5/2} + \ldots $$ You want this to be $x^{-1/2} \sin(x) \ln(x) = \left(x^{1/2} - \frac{1}{6} x^{5/2} + \ldots\right) \ln(x)$, so you get the equations $$ \eqalign{ \frac{3}{2} f_0 = 1,\ & \frac{3}{2} g_0 + f_0 = 0\cr \frac{5}{2} f_1 = 0,\ & \frac{5}{2} g_1 + f_1 = 0\cr \frac{7}{2} f_2 = \frac{-1}{6},\ & \frac{7}{2} g_2 + f_2 = 0\cr etc &}$$

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