Ytukyg

Compute
$$
I(a) = int_0^{+infty} frac {log^2(1+ax^2)}{x^4},mathrm dx,
$$
where $a > 0$.

Attempts

Easy to see that this integral converges. Also easy to see that the following integrals are convergent as well, so we could take derivatives:
$$
I'(a) = int_0^{+infty} frac {2 log(1+ax^2)}{x^2(1+ax^2)} ,mathrm dx,
$$
and
$$
frac 12 I''(a) =int_0^{+infty}frac 1{(1+ax^2)} ,mathrm dx - int_0^{+infty} frac {log(1+ax^2)}{(1+ax^2)^2},mathrm dx =: J(a) - K(a).
$$
Now for $J$, I compute as follows:
begin{align*}
J(a)&= int_{+infty}^0 frac {t^4,mathrm d(t^{-1})}{(t^2+a)^2}\
&= int_0^{+infty} frac {mathrm dt}{t^2+a} -int_0^{+infty}frac {a,mathrm dt}{(t^2+a)^2} \
&= frac pi 2cdot frac 1{sqrt a} - aint_0^{+infty} frac {mathrm dt}{(t^2+a)^2},
end{align*}
where
$$
int_0^{+infty}frac {mathrm dt}{(t^2+a)} = int_0^{+infty} -frac partial {partial a} frac 1{t^2+a} ,mathrm dt = -frac partial {partial a} left(frac pi 2 cdot frac 1{sqrt a}right) = frac pi { 4asqrt a},
$$
so
$$
J(a) = frac pi {4 sqrt a}.
$$

Difficulty

How to calculate $K(a)$ then? By using Wolfram|Alpha, I get
$$
K(a) = frac {pi(2log 2 -1)}{4sqrt a},
$$
but I do not know what I can do to reach this result. Also the indefinite integral seems awful.

Any help is appreciated, including other possible methods to compute $I$. Thanks in advance!

OK, I have solved this by myself.

Solution. $blacktriangleleft$ First we integrate by parts
begin{align*}
I(a) &= int_0^{+infty} frac 13 log^2(1 + ax^2) ,mathrm dleft(x^{-3}right) \
&= left.frac {log^2 (1 +ax^2)}{3x^3}right|_{+infty}^0 - int_{+infty}^0 frac 13 x^{-3} cdot 2log(1 + ax^2) cdot frac {2ax}{1 + ax^2}, mathrm dx \
&= int_0^{+infty} frac {4a log(1 + ax^2)}{3x^2 (1 + ax^2)} ,mathrm dx.
end{align*}

Now taking the derivatives for $I(a)$, we have
$$
I'(a) = int_0^{+infty} frac {2 log(1 + ax^2)x^2}{x^4(1 + ax^2)}, mathrm dx = int_0^{+infty} frac {2 log(1 + ax^2)}{x^2 (1+ ax^2)},mathrm dx,
$$
thus we have
$$
I(a) = frac 23 aI'(a).
$$
Now we determine the initial value. Consider computing $I'(1)$:
begin{align*}
I'(1) &= 2 int_0^{+infty} frac {log(1 + x^2)}{x^2(1+x^2)} , mathrm dx = 2int_0^{+infty} frac {log(1+ x^2 )}{x^2} ,mathrm dx - 2 int_0^{+infty} frac {log(1 +x^2)}{1+x^2} ,mathrm dx, \
int_0^{+infty} frac {log(1+ x^2)}{x^2} , mathrm dx &= int_{+infty}^0 log(1 + x^2) ,mathrm d(x^{-1})\
&= left. frac {log(1+x^2)}xright|_{+infty}^0 +int_0^{+infty} frac {2,mathrm dx}{1+ x^2} = pi.
end{align*}
For the 2nd integral, we consider another integral with a parameter
$$
K(b) = int_0^{+infty} frac {log(1 + bx^2)}{1+x^2},mathrm dx,
$$
then $K(0) = 0$. Taking the derivative, we have
begin{align*}
K'(b) &= int_0^{+infty} frac {x^2}{(1 + x^2) (1 +bx^2)} ,mathrm dx = int_0^{+infty} frac {t^{-2} ,mathrm (t^{-1})}{(1 + t^{-2}) (1+bt^{-2})}\
&= int_0^{+infty} frac 1 {b-1}left(frac 1 {1+x^2} - frac 1{b + x^2} right) , mathrm dx \
&= frac 1 {b-1} cdot frac pi 2 left(1 - frac 1{b^{1/2}}right).
end{align*}
So by the fundamental theorem of Calculus,
begin{align*}
K(1) &= K(0) + frac pi 2int_0^1 frac {sqrt b - 1}{(b-1) sqrt b} ,mathrm db = frac pi 2int_0^1 frac {mathrm db}{sqrt b(sqrt b+1)} \
&= frac pi 2 int_0^1 frac {2 ,mathrm du}{u + 1} = pi log 2.
end{align*}
Thus
$$
I'(1) = 2 times pi - 2 K(1) = 2pi (1 - log 2).
$$
Now solve the initial problem
$$
begin{cases}
I'(a) = frac 3{2a} I(a),\
I'(1) = 2pi (1 -log 2),
end{cases}
$$
we have
$$
boxed{I(a) = frac 43 a^{3/2}pi (1 - log 2)}. blacktriangleright
$$