Definite Integral $\int_{0}^{\infty} \frac{\log(x)}{(1+x+y)^{5/2}} \mathrm{d} x$

Question

My Attempt

The integral I'm trying to solve is

\begin{align*} I_1 &= \int_{0}^{\infty} \frac{\log(x)}{(1+x+y)^{5/2}} \mathrm{d} x. \end{align*}

Letting $u=\frac{1}{(1+x+y)^{3/2}}$, we have

\begin{align*} I_1 &= \frac{2}{3} \int_{0}^{(1+y)^{-3/2}} \log \left( \frac{1-u^{2/3}(1+y)}{u^{2/3}} \right) \mathrm{d} u\\ &= \frac{2}{3} \int_{0}^{(1+y)^{-3/2}} \log \left[ \frac{(1+u^{1/3}\sqrt{1+y})(1-u^{1/3}\sqrt{1+y})}{u^{2/3}} \right] \mathrm{d} u\\ &= \frac{2}{3} \biggl[ \underbrace{\int_{0}^{(1+y)^{-3/2}} \log(1+u^{1/3}\sqrt{1+y}) \mathrm{d} u}_{I_2} \, + \, \underbrace{\int_{0}^{(1+y)^{-3/2}} \log(1-u^{1/3}\sqrt{1+y}) \mathrm{d} u}_{I_3} \biggr.\\ &\qquad \biggl. - \, \frac{2}{3} \underbrace{\int_{0}^{(1+y)^{-3/2}} \log(u) \mathrm{d} u}_{I_4} \biggr]. \end{align*}

I have found that

\begin{align*} I_2 &= (1+y)^{3/2} \log(2+y) - \frac{1}{(1+y)^2} \left[ \frac{1}{3} (y+2)^3 - \frac{3}{2} (y+2)^2 + 3(y+2) - \log(y+2) - \frac{11}{6} \right] \end{align*}

and

\begin{align*} I_4 &= (1+y)^{3/2} \left[ \frac{3}{2} \log(1+y) - 1 \right]. \end{align*}

The following is my attempt at solving $I_3$. Letting $v=1-u^{1/3}\sqrt{1+y}$, we have

\begin{align} \nonumber I_3 &= \frac{1}{(1+y)^{3/2}} \int_{-y}^{1} 3(v-1)^2 \log(v) \mathrm{d} v\\ \nonumber &= \frac{1}{(1+y)^{3/2}} \left\{ \left[ \log(v) \left( v^3-3v^2+3v \right) \right]_{-y}^{1} - \int_{-y}^{1} \left( v^3 -3v^2 + 3v \right) \cdot \frac{1}{v} \mathrm{d} v \right\}\\ \nonumber &= \frac{1}{(1+y)^{3/2}} \left\{ \left[ 0-\log(-y) \left( -y^3-3y^2-3y \right) \right] - \int_{-y}^{1} \left( v^2-3v+3 \right) \mathrm{d} v \right\}\\ \nonumber &= \frac{1}{(1+y)^{3/2}} \left\{ -\log(-y) \left( -y^3-3y^2-3y \right) - \left[ \frac{1}{3}v^3 - \frac{3}{2} v^2 + 3v \right]_{-y}^{1} \right\}\\ &= \frac{1}{(1+y)^{3/2}} \left[ -\log(-y) \left( -y^3-3y^2-3y \right) - \frac{11}{6} - \frac{1}{3} y^3 - \frac{3}{2} y^2 - 3y \right]. \tag{1} \end{align}

Question

Before dealing with the main integral ($I_1$), I have proved that

\begin{align*} \int_{0}^{\infty} \frac{\log(x)}{(1+x)^{5/2}} \mathrm{d} x = \frac{4}{3} [\log(2)-1], \end{align*}

that is $I_1$ with $y=0$. Now, I would like to prove that

\begin{align*} I_1 &= \frac{2\log\left(y+1\right)+4\log\left(2\right)-4}{3\left(y+1\right)^\frac{3}{2}}. \end{align*}

In (1), I encountered a $\log(-y)$. Does this mean that I have to restrict $y$ to $y \leq 0$? How could I deal with this? Can anyone help me with this problem? I am open to any suggestions/solutions, including that for the very first integral in the question. Any help/feedback is much appreciated. Thank you!

Answer 1

Substitute $t= \frac x{1+y}$\begin{align*} &\int_{0}^{\infty} \frac{\ln x}{(1+x+y)^{5/2}} {d} x\\ =&\ \frac{1}{(1+y)^{3/2}}\int_0^\infty\frac{\ln (1+y)}{(1+t)^{5/2}}+ \frac{\ln t} {(1+t)^{5/2}}\ dt\\ =&\ \frac{2\ln(1+y)+4(\ln 2-1)}{3(1+y)^{3/2}} \end{align*} where the first integral is simply $\int_0^\infty\frac{1}{(1+t)^{5/2}}dt=\frac23$ and the second is per your $\int_{0}^{\infty} \frac{\ln t}{(1+t)^{5/2}} {d}t= \frac{4}{3} (\ln 2-1)$.

Answer 2

Define \begin{align*} I(a) &= \int_{0}^{\infty} \frac{\log(x)}{(a+x)^{3/2}} \mathrm{d} x\\ &\overset{x\to a\tan^2 (x)}=\frac2{\sqrt a}\int_0^{\pi/2}(\log(a)+2\ln\tan(x))\sin(x) \mathrm{d} x\\ &=\frac{2\log(4a)}{\sqrt a}. \end{align*} and hence $$ I'(a)=-\frac32\int_{0}^{\infty} \frac{\log(x)}{(a+x)^{5/2}} \mathrm{d} x=\frac{2-\log(4a)}{a\sqrt a}.$$ So $$ \int_{0}^{\infty} \frac{\log(x)}{(a+x)^{5/2}} \mathrm{d} x=-\frac23\frac{2-\log(4a)}{a\sqrt a}=\frac{2\log(4a)-4}{3a\sqrt a}. $$ Letting $a=1+y$ gives the desired result.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{\Large\tt I}_{1} & \equiv \color{#44f}{% \int_{0}^{\infty}{\ln\pars{x} \over \pars{1 + x + y}^{5/2}}\,\dd x} \\[5mm] & = \left.{1 \over \pars{1 + y}^{5/2}}\,\partiald{}{\nu}\int_{0}^{\infty}{x^{\color{red}{\nu}\ -\ 1} \over \bracks{\rule{0pt}{4mm}1 + x/\pars{1 + y}}^{\, 5/2}}\,\dd x\right\vert_{\nu\ =\ 1} \end{align} Note that \begin{align} & {1 \over \bracks{\rule{0pt}{4mm}1 + x/\pars{1 + y}}^{\, 5/2}} = \sum_{n = 0}^{\infty}{-5/2 \choose n}\pars{x \over 1 + y}^{n} \\[5mm] = & \ \sum_{n = 0}^{\infty}\bracks{{n + 3/2 \choose n} \pars{-1}^{n}}\pars{1 + y}^{-n}\,x^{n} \\[5mm] = & \ \sum_{n = 0}^{\infty}\color{red}{{\Gamma\pars{5/2 +n} \over \Gamma\pars{5/2}}\, \pars{1 + y}^{-n}}\,\,{\pars{-x}^{n} \over n!} \end{align} such that \begin{align} \color{#44f}{\Large\tt I}_{1} & \equiv \color{#44f}{% \int_{0}^{\infty}{\ln\pars{x} \over \pars{1 + x + y}^{5/2}}\,\dd x} \\[5mm] & = \ {1 \over \pars{1 + y}^{\, 5/2}}\ \partiald{}{\nu}\ \overbrace{\bracks{\Gamma\pars{\nu}{\Gamma\pars{5/2 - \nu} \over \Gamma\pars{5/2}}\pars{1 + y}^{\nu}}_{\nu\ =\ 1}} ^{\ds{Ramanujan's\ Master\ Theorem}} \\[5mm] & = \bbx{\color{#44f}{{2 \over 3}{\ln\pars{1 + y} + 2\ln\pars{2} - 2 \over \pars{1 + y}^{3/2}}}} \\ & \end{align}

Answer 4

Let’s consider the parametrised integral $$ I(a)=\int_0^{\infty} \frac{x^{a-1}}{(1+x+y)^{\frac{5}{2}}} d x=(1+y)^{a-\frac{5}{2}} \int_0^{\infty} \frac{u^{a-1}}{(1+u)^{\frac{5}{2}}} d u \stackrel{(*)}{=} (1+y)^{a-\frac{5}{2}} B\left(a, \frac{5}{2}-a\right) $$ where $u=\frac{x}{1+y}.$ $$ I=\left.\frac{\partial}{\partial a} \left[(1+y)^{a-\frac{5}{2}} B\left(a, \frac{5}{2}-a\right)\right]\right|_{a=1}=\frac{2[\ln (y+1)-2+2 \ln 2]}{3(y+1)^{\frac{3}{2}}} $$ For details of $(*)$, please refer to the post.