David Skrovanek
University of Pittsburgh
A Case of the Frullani Theorem
∞
∫
0
tan−1 (𝜋𝑥 ) − tan−1 (𝑥 )
𝑑𝑥
𝑥
Yipes! This looks wicked, however manageable. There is an easy way to evaluate this, using
Frullani’s Theorem, but that will come later. First let us see the unassuming solution to this
devious integral.
Think about what integrals look like when evaluated at their bounds… 𝑓 (𝑏) − 𝑓(𝑎) right? That
is somewhat reminiscent of the integrand of this problem. If we view this as the integral of
something that was just previously integrated and evaluated at its bounds, it may be more
manageable.
∞
∫
0
∞
tan−1 (𝜋𝑥 ) − tan−1 (𝑥 )
tan−1 (𝑥𝑦) 𝜋
] 𝑑𝑥
𝑑𝑥 = ∫ [
𝑥
𝑥
1
0
See it? Now this can be transformed into a double integral.
∞
𝜋
=∫ ∫
0
1
1
𝑑𝑦 𝑑𝑥
1 + 𝑥 2𝑦 2
Taking advantage of Fubini’s Theorem let’s us rewrite this as
𝜋
∞
=∫ ∫
1
0
1
𝑑𝑥 𝑑𝑦
1 + 𝑥 2𝑦 2
𝜋
tan−1 (𝑥𝑦) ∞
] 𝑑𝑦
=∫ [
𝑦
0
1
𝜋
=∫
1
=
𝜋
𝑑𝑦
2𝑦
𝜋 ln 𝜋
2
If one could not come up with this trick on their own (I certainly couldn’t), use Frullani’s
Theorem, which states
∞
∫
0
With arctan as the function,
𝑓 (𝑎𝑥 ) − 𝑓(𝑏𝑥)
𝑎
𝑑𝑥 = [𝑓 (∞) − 𝑓(0)] ln ( )
𝑥
𝑏
David Skrovanek
University of Pittsburgh
∞
tan−1 (𝜋𝑥 ) − tan−1 (𝑥 )
𝜋
𝜋
∫
𝑑𝑥 = [ − 0] ln ( )
𝑥
2
1
0
=
𝜋 ln 𝜋
2
Using that same idea, we can evaluate some integrals that may be a little less obvious.
1
∫
0
𝑥−1
𝑑𝑥
ln(𝑥)
Okay, so maybe this looks nothing like the general case of Frullani’s Integral, but it can be
adjusted. Consider this,
∞
𝑒 −𝑎𝑥 − 𝑒 −𝑏𝑥
∫
𝑑𝑥
𝑥
0
So this case is pretty obvious; it is simply equal to ln(𝑏) − ln(𝑎).
𝐿𝑒𝑡 𝑢 = 𝑒 −𝑥 ,
𝑑𝑢 = −𝑒 𝑥 𝑑𝑥 →
𝑑𝑥 =
−𝑑𝑢
𝑢
Notice how the bounds change too:
0
0 𝑎−1
1 𝑏−1
−𝑑𝑢 𝑢𝑎 − 𝑢𝑏
𝑢
− 𝑢𝑏−1
𝑢
− 𝑢𝑎−1
∫
(
)=∫
𝑑𝑢 = ∫
𝑑𝑢 = ln(𝑏) − ln(𝑎)
𝑢
−ln(𝑢)
ln(𝑢)
ln(𝑢)
1
1
0
Looking better yet? Getting back to the original problem, we observe that 𝑥 − 1 is the numerator
of the derived expression above when 𝑎 = 1 and 𝑏 = 2. Thus, we can now see that
1
∫
0
A very special case indeed!
𝑥−1
𝑑𝑥 = ln(2)
ln(𝑥)