EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 3, Article Number 6575
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On the Evaluation of Certain Unsolved Definite
Integrals

Irshad Ayoob
1 Department of Mathematics and Sciences, Prince Sultan University, P.O. Box 66833,
Riyadh 11586, Saudi Arabia

Abstract. We study the following three definite integrals, previously posed as open problems by
another researcher: I(α) =

∫∞
0

x−1/2 ln(1 + x−α) dx, In(α) =
∫∞
0

1√
x(x2+4α2)n

dx, and I(β) =∫∞
0

x−3/2
[
f(2β/x) − f(2/x)

]
dx. We establish sufficient conditions for the convergence of these

integrals and evaluate them in closed form using special functions. In particular, the third integral
I(β) turns out to be similar to Frullani integral, and we obtain two interesting formulas for this
integral. These types of integrals have been used to establish logarithmic Hardy-Hilbert-type
inequalities.

2020 Mathematics Subject Classifications: 26A42, 33B15, 26A06, 44A20

Key Words and Phrases: Definite integral, Beta function, Leibniz integral rule, Mellin transform

Definite integrals appear in various contexts across both pure and applied branches of
mathematics. While many definite integrals can be solved with elementary techniques such
as substitution or integration by parts, a significant number of them are non-elementary
that cannot be expressed in terms of basic functions. These integrals demand more ad-
vanced methods for exact evaluation, including transformations (e.g., Laplace or Mellin),
complex analysis (e.g., contour integration and residue theory), and special functions (such
as the Gamma, Beta, and hypergeometric functions).

Finding exact solutions, when possible, is of high practical and theoretical impor-
tance as it allows for precise predictions, deeper analytic understanding, and verification
of numerical methods. As such, the study of advanced techniques for evaluating definite
integrals remains an active area of research. An extensive compilation of definite integrals,
ranging from elementary to non-elementary forms, is available in [1] and the references
cited therein. Recent studies, such as those presented in [2–7], highlight ongoing develop-
ments and underscore the sustained interest in this field.

The evaluation of the following definite integrals are posed as open problems in [8].

DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6575

Email address: iayoub@psu.edu.sa (I. Ayoob)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 2 of 12

Problem 1. Evaluate the integral

I(α) =

∫ ∞

0
x−1/2 ln

(
1 + x−α

)
dx, for α > 1

2 .

Problem 2. Determine a closed-form expression for

In(α) =

∫ ∞

0

1√
x (x2 + 4α2)n

dx, for α > 0, n ∈ N.

Problem 3. Evaluate

I(β) =

∫ ∞

0
x−3/2

[
f

(
2β

x

)
− f

(
2

x

)]
dx, for a suitable function f and β > 0.

The author in [8] has evaluated the special cases of the mentioned integrals. We list
the special cases as following.

Corollary 1. Proposition 2.8 in [8] is a special case of Problem 1 with α = 2.

Corollary 2. Proposition 2.2, 2.3, and 2.4 in [8] are special cases of Problem 2 with
n = 1, 2 and n = 3 respectively.

Corollary 3. Proposition 2.5 in [8] is a special case of Problem 3 with f(x) = arctan(x).

The further implications of these corollaries yield some integral formulas for π (see [8]).
As a further applications, the author in [8] has applied the special cases of the integrals
given in Problem 1,2 and 3 to obtain the logarithmic Hardy-Hilbert-type inequalities [9]
(for example see Proposition 2.11 etc).

Our goal is to establish the convergence results for the integrals given in Problem 1, 2
and 3, and evaluate them in closed form for general parameters n, α and β.

1. Main Results

1.1. Convergence and evaluation of the first integral.

Here is the convergence result for the first integral.

Proposition 1. Let

I(α) =

∫ ∞

0
x−

1
2 ln

(
1 + x−α

)
dx.

If α > 1
2 , then the improper integral I(α) converges.

Proof. We split the integral at x = 1,

I(α) =

∫ 1

0
x−

1
2 ln

(
1 + x−α

)
dx +

∫ ∞

1
x−

1
2 ln

(
1 + x−α

)
dx = I1 + I2.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 3 of 12

(i) Convergence of I2: For x ≥ 1, 0 ≤ x−α ≤ 1, and since ln(1 + u) ≤ u for u ≥ 0, we
have

0 ≤ ln
(
1 + x−α

)
≤ x−α.

Hence

0 ≤ I2 ≤
∫ ∞

1
x−

1
2x−α dx =

∫ ∞

1
x−(α+

1
2) dx.

Because α+ 1
2 > 1, this last integral converges, so I2 < ∞.

(ii) Convergence of I1: On (0, 1], x−α ≥ 1. For u ≥ 1, ln(1 + u) = lnu+ ln(1 + u−1) ≤
lnu+ ln 2. Setting u = x−α gives

ln
(
1 + x−α

)
≤ ln 2 + α(− lnx).

Therefore for 0 < x ≤ 1,

0 ≤ x−
1
2 ln

(
1 + x−α

)
≤ (ln 2)x−

1
2 + αx−

1
2 (− lnx).

We check each term: ∫ 1

0
x−

1
2 dx = 2,

∫ 1

0
x−

1
2 (− lnx) dx = 4.

Hence
I1 ≤ (ln 2) · 2 + α · 4 < ∞.

Combining (i) and (ii) shows I1 < ∞ and I2 < ∞. Thus I(α) converges for all α > 1
2 .

Now we solve the first integral.

Theorem 1. For every α > 1
2 , we have

I(α) =

∫ ∞

0
x−

1
2 ln

(
1 + x−α

)
dx = 2π csc

(
π
2α

)
.

Proof. We write

I(α) =

∫ ∞

0
x−

1
2 ln

(
1 + x−α

)
dx.

Since
∂

∂α
ln
(
1 + x−α

)
= −x−α lnx

1 + x−α
,

we may differentiate under the integral sign to obtain

dI

dα
= −

∫ ∞

0
x−

1
2
x−α lnx

1 + x−α
dx = −

∫ ∞

0

x−
1
2 lnx

xα + 1
dx.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 4 of 12

Set u = xα, so that

x = u1/α, dx =
1

α
u

1
α−1du, lnx =

1

α
lnu, x−

1
2 = u−

1
2α .

Then

dI

dα
= − 1

α2

∫ ∞

0

u−
1
2α+

1
α−1 lnu

u+ 1
du = − 1

α2

∫ ∞

0

us−1 lnu

1 + u
du,

where s = 1
2α ∈ (0, 1). The following integral is well-known (see [1]),∫ ∞

0

us−1

1 + u
du =

π

sin(πs)
,

and differentiating in s yields∫ ∞

0

us−1 lnu

1 + u
du = −π2 cos(πs)

sin2(πs)
.

Hence
dI

dα
=

π2

α2

cos
(

π
2α

)
sin2

(
π
2α

) .
Next, we set t = π

2α , so that α = π
2t and

dα = − π

2t2
dt,

π2

α2
= 4t2.

Thus
dI

dα
dα = 4t2

cos t

sin2 t

(
− π

2t2
dt
)
= −2π

cos t

sin2 t
dt = 2π d

(
csc t

)
.

Integrating shows

I(α) = 2π csc
(

π
2α

)
+ C.

Finally,

I(2) = 2π csc
(

π
2·2

)
+ C = 2π csc

(
π
4

)
+ C = 2π

√
2 + C.

Since we know [8] that I(2) = 2π
√
2, it follows that

2π
√
2 = 2π

√
2 + C =⇒ C = 0.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 5 of 12

1.2. Convergence and evaluation of the second integral.

Here is the convergence result for the second integral.

Proposition 2. Let α > 0 and n ∈ N, n ≥ 1. Then

In(α) =

∫ ∞

0

1
√
x
(
x2 + 4α2

)n dx

converges.

Proof. We split the integral at x = 1,

In(α) =

∫ 1

0

dx√
x (x2 + 4α2)n

+

∫ ∞

1

dx√
x (x2 + 4α2)n

= I1 + I2.

1. Convergence on [0, 1] :
For 0 < x ≤ 1, we have

x2 + 4α2 ≥ 4α2,

so
1√

x (x2 + 4α2)n
≤ 1√

x (4α2)n
= (4α2)−n x−

1
2 .

Since ∫ 1

0
x−

1
2 dx = 2 < ∞,

it follows by the comparison test that I1 < ∞.

2. Convergence on [1,∞) :
For x ≥ 1, we have

x2 + 4α2 ≥ x2,

so
1√

x (x2 + 4α2)n
≤ 1√

xx2n
= x−

(
2n+

1
2

)
.

Since 2n+ 1
2 > 1, the p-integral

∫ ∞

1
x− (2n+

1
2)dx converges. Hence I2 < ∞ by comparison.

Combining these two estimates shows In(α) = I1 + I2 < ∞, as claimed.

Next, we solve the second integral.

Theorem 2. For every α > 0 and integer n ≥ 1, we have the following integral

In(α) =

∫ ∞

0

dx√
x (x2 + 4α2)n

=
1

2
(4α2)−

(
n−1

4

)
Γ
(
1
4

)
Γ
(
n− 1

4

)
Γ(n)

.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 6 of 12

Proof.
We set

x = 2α t, dx = 2αdt,
√
x = (2α)

1
2 t

1
2 , x2 + 4α2 = 4α2(1 + t2).

Therefore

In(α) =

∫ ∞

0

1

(2α)
1
2 t

1
2 [4α2(1 + t2)]n

(2αdt) = (2α)
1
2 (4α2)−n

∫ ∞

0

dt

t
1
2 (1 + t2)n

.

Since

(2α)
1
2 (4α2)−n = 2

1
2−2n α

1
2−2n,

we have

In(α) = 2
1
2−2n α

1
2−2n

∫ ∞

0
t−

1
2 (1 + t2)−n dt.

Set

Jn =

∫ ∞

0
t−

1
2 (1 + t2)−n dt.

With the substitution u = t2, du = 2t dt, t−
1
2 = u−

1
4 , we get

Jn =

∫ ∞

0
u−

1
4 (1 + u)−n du

2u
1
2

=
1

2

∫ ∞

0
u
1
4−1(1 + u)−n du =

1

2
B
(
1
4 , n− 1

4

)
.

For s > 0, recall the definition of gamma function Γ(s) =
∫∞
0 ts−1e−t dt, and the relation-

ship with the beta function, B(x, y) = Γ(x)Γ(y)/Γ(x+ y), it follows that

Jn =
1

2

Γ(14) Γ(n− 1
4)

Γ(n)
.

Hence

In(α) = 2
1
2−2n α

1
2−2n 1

2

Γ(14) Γ(n− 1
4)

Γ(n)
=

1

2
(4α2)−(n−

1
4 ) Γ(14) Γ(n− 1

4)

Γ(n)
,

as claimed.

1.3. Convergence and evaluation of the third integral.

Here is the convergence result for the third integral.

Proposition 3. Let f : (0,∞) → R be a continuously differentiable function such that

lim
t→0+

f(t) = f(0) ∈ R, lim
t→∞

f(t) = f(∞) ∈ R,



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 7 of 12

and ∫ ∞

0
t
1
2
∣∣f ′(t)

∣∣ dt < ∞.

Then for each fixed β > 0 the integral

I(β) =

∫ ∞

0
x−

3
2

[
f
(2β

x

)
− f

(
2
x

)]
dx

converges absolutely.

Proof. Fix β > 0. We may write

I(β) =

∫ ∞

0
x−3/2

[
f(2β/x)−f(2/x)

]
dx =

∫ x0

0
+

∫ X1

x0

+

∫ ∞

X1

x−3/2
[
f(2β/x)−f(2/x)

]
dx,

where we choose 0 < x0 < X1 < ∞ so that

2β

x0
≥ T,

2

X1
≤ S,

with T large and S small enough that
∫∞
T t

1
2 |f ′(t)| dt < ε and

∫ S
0 t

1
2 |f ′(t)| dt < ε.

(i) Tail as x → 0+: For 0 < x ≤ x0, both
2β
x and 2

x lie in [T,∞). By the Mean-Value

Theorem there exists ξ ∈ [ 2x ,
2β
x ] such that

f
(2β

x

)
− f

(
2
x

)
= f ′(ξ)

(
2β
x − 2

x

)
= (β − 1)

2

x
f ′(ξ).

Hence ∣∣x−3/2[ f(2β/x)− f(2/x)]
∣∣ = 2|β − 1| x−5/2

∣∣f ′(ξ)
∣∣.

Since ξ ≥ T and t 7→ t1/2|f ′(t)| is integrable on [T,∞), the change of variable ξ = 2θ/x
shows ∫ x0

0
x−3/2

∣∣f(2β/x)− f(2/x)
∣∣ dx = 2|β − 1|

∫ ∞

T
t
1
2
∣∣f ′(t)

∣∣ dt < 2|β − 1| ε.

(ii) Tail as x → ∞: For x ≥ X1, both
2β
x and 2

x lie in (0, S]. Again by the Mean-Value

Theorem there is η ∈ [ 2x ,
2β
x ] ⊂ (0, S] such that

f
(2β

x

)
− f

(
2
x

)
= f ′(η)

(
2β
x − 2

x

)
= (β − 1)

2

x
f ′(η),

and hence∫ ∞

X1

x−3/2
∣∣f(2β/x)− f(2/x)

∣∣ dx = 2|β − 1|
∫ S

0
t
1
2
∣∣f ′(t)

∣∣ dt < 2|β − 1| ε.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 8 of 12

(iii) Middle region: On the compact interval [x0, X1], the function

x 7→ x−3/2
[
f(2β/x)− f(2/x)

]
is continuous and hence bounded, so its integral is finite.

Combining (i), (ii), and (iii) shows I(β) =

∫ ∞

0
x−

3
2

[
f
(2β

x

)
− f

(
2
x

)]
dx < ∞. Therefore

I(β) converges absolutely.

We have the following result which is similar to Frullani type integral.

Theorem 3. Let f : (0,∞) → R be a continuously differentiable function such that

lim
t→0+

f(t) = f(0) ∈ R, lim
t→∞

f(t) = f(∞) ∈ R,

and

A =

∫ ∞

0
t
1
2
∣∣f ′(t)

∣∣ dt < ∞.

Then for every β > 0,∫ ∞

0
x−

3
2

[
f
(2β

x

)
− f

(
2
x

)]
dx =

√
2A

(
1− β−1

2
)
.

Proof. Define

I(β) =

∫ ∞

0
x−

3
2

[
f
(2β

x

)
− f

(
2
x

)]
dx.

Since f is C1 and the subtraction makes the integrand absolutely convergent at both ends,
we may differentiate under the integral sign:

dI

dβ
=

∫ ∞

0
x−

3
2

∂

∂β

[
f
(2β

x

)]
dx = 2

∫ ∞

0
x−

5
2 f ′(2β

x

)
dx.

Perform the substitution t = 2β
x , so x = 2β

t and dx = − 2β
t2

dt. Then

x−
5
2 =

(
2β
t

)− 5
2
= (2β)−

5
2 t

5
2 ,

and
x−

5
2 dx = (2β)−

5
2 t

5
2
(
−2β

t2

)
dt = −(2β)−

3
2 t

1
2 dt.

Hence∫ ∞

0
x−

5
2 f ′(2β

x

)
dx = (2β)−

3
2

∫ 0

∞
t
1
2 f ′(t) (−dt) = (2β)−

3
2

∫ ∞

0
t
1
2 f ′(t) dt = (2β)−

3
2 A,

where A =
∫∞
0 t

1
2
∣∣f ′(t)

∣∣ dt < ∞.
Thus

dI

dβ
= 2 (2β)−

3
2 A = 2−

1
2 Aβ− 3

2 .



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 9 of 12

Integrating from β = 1 (where clearly I(1) = 0) to a general β > 0 gives

I(β) =

∫ β

1

dI

db
db = 2−

1
2 A

∫ β

1
b−

3
2 db = 2−

1
2 A

[
−2 b−

1
2
]β
1
=

√
2A

(
1− β−1

2
)
.

This proves the general formula.

The following special case in [8] is deduced from Theorem 3.

Remark 1. In the special case, f(x) = arctanx, we have f ′(t) = 1/(1 + t2), and the
constant

A =

∫ ∞

0

t
1
2

1 + t2
dt = 1

2 B
(
3
4 ,

1
4

)
=

π
√
2

2
,

yielding ∫ ∞

0

arctan
(2β

x

)
− arctan

(
2
x

)
x
√
x

dx = π
(
1− β−1

2
)
.

Note that Theorem 3 requires f to be continuously differentiable. The following result
gives one more closed form solution of the integral I(β) for more general class of functions
in terms of Mellin transform.

Proposition 4. Let f : (0,∞) → R be measurable and suppose its Mellin transform

Mf (s) =

∫ ∞

0
ts−1f(t) dt

converges at s = 1
2 , i.e. ∫ ∞

0
t−1/2

∣∣f(t)∣∣ dt < ∞.

Then for each fixed β > 0 the integral

I(β) =

∫ ∞

0

1

x
√
x

[
f
(2β

x

)
− f

(
2
x

)]
dx

converges absolutely.

Proof. We set

g(x) =
1

x
√
x

[
f
(2β

x

)
− f

(
2
x

)]
.

Make the substitution

t =
2

x
, x =

2

t
, dx = − 2

t2
dt.

Then
dx

x
√
x
=

−2

t2
1

(2/t)
√
2/t

= −2−
1
2 t−

1
2 dt,

and
2β

x
= βt,

2

x
= t.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 10 of 12

Hence

I(β) =

∫ ∞

0
g(x) dx =

∫ 0

∞
f(βt)− f(t) (−2−

1
2 t−

1
2 ) dt

= 2−
1
2

∫ ∞

0
t−

1
2
[
f(βt)− f(t)

]
dt.

It suffices to show ∫ ∞

0
t−

1
2
∣∣f(βt)− f(t)

∣∣ dt < ∞.

By the triangle inequality, ∣∣f(βt)− f(t)
∣∣ ≤ |f(βt)|+ |f(t)|,

so ∫ ∞

0
t−

1
2
∣∣f(βt)− f(t)

∣∣ dt ≤ ∫ ∞

0
t−

1
2 |f(βt)| dt+

∫ ∞

0
t−

1
2 |f(t)| dt.

For the first term, substitute u = βt, du = β dt:∫ ∞

0
t−

1
2 |f(βt)| dt = β−1

2

∫ ∞

0
u−

1
2 |f(u)| du.

Hence ∫ ∞

0
t−

1
2
(
|f(βt)|+ |f(t)|

)
dt =

(
1 + β−1

2
) ∫ ∞

0
t−

1
2 |f(t)| dt,

which is finite by hypothesis. Therefore I(β) converges absolutely.

Theorem 4. Let f : (0,∞) → R be measurable and suppose its Mellin transform

Mf (s) =

∫ ∞

0
ts−1f(t) dt

converges at s = 1
2 , i.e. ∫ ∞

0
t−1/2

∣∣f(t)∣∣ dt < ∞.

Then we have,

I(β) =

∫ ∞

0

1

x
√
x

[
f
(2β

x

)
− f

(
2
x

)]
dx =

1√
2

(
β−1/2 − 1

)
Mf

(
1
2

)
.

Proof.
Starting from the following convergent integral (Proposition 4 ensures convergence),

I(β) =

∫ ∞

0

1

x
√
x

[
f(2β/x)− f(2/x)

]
dx,



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 11 of 12

we perform the change of variable t = 2/x, so that x = 2/t and dx = −2 t−2dt. As in the
proof of Proposition 4,

dx

x
√
x
= −2−

1
2 t−

1
2 dt,

2β

x
= βt,

2

x
= t.

Hence

I(β) =

∫ 0

∞

[
f(βt)− f(t)

] (
−2−

1
2 t−

1
2
)
dt = 2−

1
2

∫ ∞

0
t−

1
2
[
f(βt)− f(t)

]
dt

= 2−
1
2

[∫ ∞

0
t−

1
2 f(βt) dt−

∫ ∞

0
t−

1
2 f(t) dt

]
.

In the first integral substitute u = βt, du = β dt, giving∫ ∞

0
t−

1
2 f(βt) dt = β−1

2

∫ ∞

0
u−

1
2 f(u) du = β−1

2 Mf

(
1
2

)
.

The second integral is exactly Mf (
1
2). Therefore

I(β) = 2−
1
2

[
β−1

2Mf (
1
2) − Mf (

1
2)
]
=

Mf (
1
2)√
2

(
β−1

2 − 1
)
,

as claimed.

2. Conclusion

In this work, we have evaluated three classes of definite integrals that were originally
posed as open problems in the literature. Our approach involved rigorously establishing
sufficient conditions under which these integrals converge, followed by the derivation of
explicit closed-form expressions. These results are presented in Theorems 1, 2, 3 and 4 of
the paper. Beyond their intrinsic analytical interest, these integrals serve as foundational
kernels in constructing a new class of generalized logarithmic Hardy–Hilbert-type inequal-
ities, extending those previously established in [8]. The identification of optimal constants,
further generalization to multidimensional or operator-theoretic settings, and exploration
of applications in functional inequalities represent promising directions for future research.

Acknowledgements

The author would like to thank the Prince Sultan University for paying the publication
fees for this work through TAS LAB.



I. Ayoob / Eur. J. Pure Appl. Math, 18 (3) (2025), 6575 12 of 12

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