I would like to solve the following parameterised definite integral. $$ f(a) = \int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx $$ Where $a \in \mathbb{R}^+$
I have tried a few different methods that haven't resulted in anything useful. I was hoping to ask for a starting point on this question.
Note - if possible, I'm trying to resolve this integral using Real Analysis only.
As per a recommendation provided, I will employ Feynman's Trick (coupled with the Dominated Convergence Theorem and Leibniz's Integral Rule) and differentiate. Firstly though, we observe that $f(0) = 0$. We now proceed by differentiating under the curve with respect to $a$:
\begin{align} f'(a)& = \frac{d}{da}\int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx = \int_0^1 \frac{\frac{\partial }{\partial a}\big[\ln\left(ax^2 + 1\right)\big]}{x + 1}\:dx \\ & = \int_0^1 \frac{x^2}{ax^2 + 1} \cdot \frac{1}{1 + x}\:dx = \int_0^1 \frac{x^2}{\left(ax^2 + 1\right)\left(x + 1\right)}\:dx \end{align} We now apply a Partial Fraction Decomposition on the integrand: \begin{align} f'(a) &= \int_0^1 \frac{1}{a + 1}\left[\frac{x}{ax^2 + 1} - \frac{1}{ax^2 + 1} + \frac{1}{x + 1}\right]\:dx\\ & = \frac{1}{a + 1}\left[\frac{1}{2a}\ln\left(ax^2 + 1\right) - \frac{1}{\sqrt{a}}\arctan\left( \sqrt{a}x\right) + \ln\left(x + 1\right) \right]_0^1 \\ &= \frac{1}{a + 1}\left[\frac{\ln\left(a + 1\right)}{2a} - \frac{\arctan\left( \sqrt{a}\right) }{\sqrt{a}} + \ln(2) \right] \end{align}
From here, we resolve $f(a)$ by integrating with respect to $a$: \begin{align} f(a) &= \int \frac{1}{a + 1}\left[\frac{\ln\left(a + 1\right)}{2a} - \frac{\arctan\left( \sqrt{a}\right) }{\sqrt{a}} + \ln(2) \right]\:da \\ &= \frac{1}{2}\int \frac{\ln(a + 1)}{a\left(a + 1\right)}\:da - \int \frac{\arctan\left(\sqrt{a}\right)}{\sqrt{a}\left(a + 1\right)} + \ln(2)\int \frac{1}{a + 1}\:da \\ &= \frac{1}{2}A - B + \ln(2)D \end{align} We now resolve each individually (note I will omit the constant of integration till the end). Starting with the easiest $D$: $$ D = \int \frac{1}{a + 1}\:da = \ln(a + 1) $$ We now resolve $A$ by applying a Partial Fraction Decomposition: \begin{align} A &= \int \ln(a + 1) \left[ \frac{1}{a} - \frac{1}{a + 1} \right]\:da = \int \frac{\ln(a + 1)}{a}\:da - \int \frac{\ln(a + 1)}{a + 1}\:da \\ &= -\operatorname{Li}_2(-a) - \frac{1}{2}\ln^2(a + 1) \end{align} We now resolve $B$. Here we make the substitution $a = b^2$, $b > 0$ \begin{align} B = \int \frac{\arctan(b)}{b\left(b^2 + 1\right)} \cdot 2b \:db = 2\int \frac{\arctan(b)}{b^2 + 1}\:db = 2 \cdot \frac{1}{2}\arctan^2(b) = \arctan^2\left(\sqrt{a}\right) \end{align} Thus, we now may form $f(a)$ \begin{align} f(a) &= \frac{1}{2}A - B + \ln(2)D \\ &=\frac{1}{2}\left[-\operatorname{Li}_2(-a) - \frac{1}{2}\ln^2(a + 1) \right] - \arctan^2\left(\sqrt{a}\right) + \ln(2)\ln(a + 1) + C \end{align} Where $C$ is the constant of integration. We resolve $C$ using $f(0) = 0$: \begin{align} f(0) = 0 = \frac{1}{2}\left[-\operatorname{Li}_2(0) - \frac{1}{2}\ln^2(1) \right] - \arctan^2\left(\sqrt{0}\right) + \ln(2)\ln(0 + 1) + C = 0 + C \rightarrow C = 0 \end{align} And so, $$ \int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx = \frac{1}{2}\left[-\operatorname{Li}_2(-a) - \frac{1}{2}\ln^2(a + 1) \right] - \arctan^2\left(\sqrt{a}\right) + \ln(2)\ln(a + 1) $$
$$x = \frac{1- t}{1 + t}$$
As well as using that under the region of integration we can use
$$ \frac{1}{1 + x} = \sum_{n = 0}^\infty (-1)^n x^n $$
For both (in my attempts) it only resulted in very messy algebra.
– David Galea Aug 07 '20 at 04:46