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!
Best Answer
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 $\int_{0}^{\infty} \frac{\ln t}{(1+t)^{5/2}} {d}t= \frac{4}{3} (\ln 2-1)$.