American Journal of Interdisciplinary Research and Development 

ISSN Online: 2771-8948 

Website: www.ajird.journalspark.org 

Volume 11, Dec., 2022 

35 | P a g e  
 

 

ASYMPTOTIC EVALUATION OF PARAMETER-

DEPENDENT INTEGRALS 
Umirzakova Iroda  

3rd Year Student of the Faculty of Mathematics of Samarkand State University 

named after Sharof Rashidov 

 

Komilov Abdulaziz  

 3rd Year Student of the Faculty of Mathematics of Samarkand State University 

named after Sharof Rashidov 

 

Abstract 

This thesis presents the methods of calculating integrals that are approximate, but 

the calculation of their value is quite complicated. 

 

Keywords: parameter-dependent integral, asymptotic estimation, Taylor series, 

Euler integrals. 
 

 

When constructing mathematical models of life problems, in most cases, it is important to know 

the approximate value of the solution closest to this solution, rather than finding the exact value 

or values of the problem solution. This thesis considers the issue of approximate calculation of 

integrals depending on the parameter or asymptotic estimation of the parameter, which plays 

an important role in solving this type of problems. 

In mathematics, an analytic function is a function that is locally given by a convergent power 

series. 

First, we present the following theorem: 

Theorem. If every term of the series ∑ 𝑢𝑛(𝑥)∞
𝑛=1  is continuous in the segment [𝑎, 𝑏], and this 

series is uniformly convergent in this segment and 𝑆(𝑥) is a sum of series, then the following 

equality holds: 

∫ 𝑆(𝑥)
𝑏

𝑎
𝑑𝑥 = ∑ ∫ 𝑢𝑛(𝑥)

𝑏

𝑎
𝑑𝑥∞

𝑛=1 . 

 

Let the following integral depending on parameters 𝑝 and 𝑞 be given:  

              Φ(𝑝, 𝑞) = ∫ 𝑓(𝑥, 𝑝) ∙ 𝑔(𝑥, 𝑞)
𝑏

𝑎
𝑑𝑥                                    (1)   

Let us assume that the functions 𝑓(𝑥, 𝑝), 𝑔(𝑥, 𝑞) are analytic in some interval (−𝜌, 𝜌), and this 

condition (𝑎, 𝑏) ⊂ (−𝜌, 𝜌) is hold. Then the following equations hold for ∀𝑥𝜖(𝑎, 𝑏):  

https://en.wikipedia.org/wiki/Mathematics
https://en.wikipedia.org/wiki/Function_(mathematics)
https://en.wikipedia.org/wiki/Convergent_series
https://en.wikipedia.org/wiki/Power_series
https://en.wikipedia.org/wiki/Power_series


 

American Journal of Interdisciplinary Research and Development 

ISSN Online: 2771-8948 

Website: www.ajird.journalspark.org 

Volume 11, Dec., 2022 

36 | P a g e  
 

 

𝑓(𝑥, 𝑝) = ∑
𝑓𝑛(0, 𝑝)

𝑛!
∙ 𝑥𝑛

∞

𝑛=0

    𝑔(𝑥, 𝑞) = ∑
𝑓𝑛(0, 𝑞)

𝑛!
∙ 𝑥𝑛

∞

𝑛=0

 

Using these equations and the above theorem, we can write integral (1) in the following form: 

Φ(𝑝, 𝑞) = ∫ ∑
𝑓𝑛(0, 𝑝)

𝑛!
∙ 𝑥𝑛 ∙ 𝑔(𝑥, 𝑞)

∞

𝑛=0

𝑏

𝑎

𝑑𝑥 = ∑
𝑓𝑛(0, 𝑝)

𝑛!
∫𝑔(𝑥, 𝑞) ∙ 𝑥𝑛

𝑏

𝑎

𝑑𝑥

∞

𝑛=0

, 

Φ(𝑝, 𝑞) = ∫ ∑
𝑔𝑛(0, 𝑞)

𝑛!
∙ 𝑥𝑛 ∙ 𝑓(𝑥, 𝑝)

∞

𝑛=0

𝑏

𝑎

𝑑𝑥 = ∑
𝑔𝑛(0, 𝑞)

𝑛!
∫ 𝑓(𝑥, 𝑝) ∙ 𝑥𝑛

𝑏

𝑎

𝑑𝑥

∞

𝑛=0

. 

Here 
𝑓𝑛(0,𝑝)

𝑛!
 and 

𝑔𝑛(0,𝑞)

𝑛!
  are the Taylor coefficients of the functions 𝑓(𝑥, 𝑝) and 𝑔(𝑥, 𝑞), 

respectively. 

For convenience, we choose one of the integrals ∫ 𝑥𝑛𝑓(𝑥, 𝑝)
𝑏

𝑎
𝑑𝑥 and  ∫ 𝑥𝑛𝑔(𝑥, 𝑞)

𝑏

𝑎
𝑑𝑥 in such 

a way that to calculate the chosen integral be easy. Suppose that the equality∫ 𝑥𝑛𝑓(𝑥, 𝑝)
𝑏

𝑎
𝑑𝑥 =

𝜓(𝑛, 𝑝) is hold, then 

∫𝑓(𝑥, 𝑝) ∙ 𝑔(𝑥, 𝑞)

𝑏

𝑎

𝑑𝑥 = ∫ ∑
𝑔𝑛(0, 𝑞)

𝑛!
∙ 𝑥𝑛𝑓(𝑥, 𝑝)

∞

𝑛=0

𝑏

𝑎

𝑑𝑥 = 

= ∑
𝑔𝑛(0, 𝑞)

𝑛!
∫𝑓(𝑥, 𝑝) ∙ 𝑥𝑛

𝑏

𝑎

𝑑𝑥

∞

𝑛=0

= ∑
𝑔𝑛(0, 𝑞)

𝑛!
∙ 𝜓(𝑛, 𝑝)

∞

𝑛=0

. 

If we choose 𝑛 < 𝑛0 in this equation, we arrive at the following approximate equation: 

∫ 𝑓(𝑥, 𝑝) ∙ 𝑔(𝑥, 𝑞)

𝑏

𝑎

𝑑𝑥 ≈ ∑
𝑔𝑛(0, 𝑞)

𝑛!
∙ 𝜓(𝑛, 𝑝)

𝑛0

𝑛=0

 

It is known that the larger the number 𝑛0 can be, the closer the found value is to the value of 

the given integral. 

Some methods of calculating integrals depending on parameter are given below: 

Problem 1. Evaluate the value of the following integral:   𝐼(𝑟) = ∫ 𝑒𝑚𝑟∙cos𝑥𝜋

0
𝑑𝑥. 

Solution. First, we split this integral into two integrals, and then get the substitution:  

𝐼(𝑟) = ∫ 𝑒𝑚𝑟∙cos𝑥

𝜋

0

𝑑𝑥 = ∫ 𝑒𝑚𝑟∙cos𝑥

𝜋
2

0

𝑑𝑥 + ∫ 𝑒𝑚𝑟∙cos𝑥

𝜋

𝜋
2

𝑑𝑥 =

[
 
 
 
 
 
 𝑥 = 𝑡 +

𝜋

2
𝑑𝑥 = 𝑑𝑡

𝑥 =
𝜋

2
→ 𝑡 = 0

𝑥 = 𝜋 → 𝑡 =
𝜋

2]
 
 
 
 
 
 

= 



 

American Journal of Interdisciplinary Research and Development 

ISSN Online: 2771-8948 

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Volume 11, Dec., 2022 

37 | P a g e  
 

 

= ∫ 𝑒𝑚𝑟∙cos𝑥

𝜋
2

0

𝑑𝑥 + ∫ 𝑒𝑚𝑟∙cos(𝑡+
𝜋
2
)

𝜋
2

0

𝑑𝑡 = ∫ 𝑒𝑚𝑟∙cos𝑥

𝜋
2

0

𝑑𝑥 + ∫ 𝑒−𝑚𝑟∙sin𝑥

𝜋
2

0

𝑑𝑥. 

We can substitute in the integrals on the right side of the last equality: 

∫ 𝑒𝑚𝑟∙cos𝑥

𝜋
2

0

𝑑𝑥 = [cos 𝑥 = 𝑡] = ∫
𝑒𝑚𝑟𝑡

√1 − 𝑡2
𝑑𝑡

1

0

, 

∫ 𝑒−𝑚𝑟∙sin𝑥

𝜋
2

0

𝑑𝑥 = [sin 𝑥 = 𝑧] = ∫
𝑒−𝑚𝑟𝑧

√1 − 𝑧2
𝑑𝑧

1

0

= ∫
𝑒−𝑚𝑟𝑡

√1 − 𝑡2
𝑑𝑡

1

0

. 

Then,  

𝐼(𝑟) = ∫
𝑒𝑚𝑟𝑡 + 𝑒−𝑚𝑟𝑡

√1 − 𝑡2
𝑑𝑡

1

0

. 

Now we use this expansion: 𝑒𝑥 = ∑
𝑥𝑛

𝑛!
∞
𝑛=0 . Then we arrive at the following equality: 

𝐼(𝑟) = ∫(
1

√1 − 𝑡2
∑

(𝑚𝑟𝑡)𝑛 + (−𝑚𝑟𝑡)𝑛

𝑛!

∞

𝑛=0

)

1

0

𝑑𝑡. 

It is known that in the sum on the right side of the above equation, the odd-numbered terms 

become zero. So, 

𝐼(𝑟) = 2∫ ∑
(𝑚 ∙ 𝑟)2𝑛𝑡2𝑛

2𝑛! ∙ √1 − 𝑡2

∞

𝑛=0

1

0

𝑑𝑡 = ∑
(𝑚𝑟)2𝑛

(2𝑛)!
∫

2𝑡2𝑛

√1 − 𝑡2

1

0

𝑑𝑡

∞

𝑛=0

=

[
 
 
 

𝑡2 = 𝑥
𝑡 = √𝑥

𝑑𝑡 =
1

2√𝑥
𝑑𝑥

]
 
 
 

= 

= 2 ∑
(𝑚𝑟)2𝑛

(2𝑛)!
∫

𝑥𝑛

(1 − 𝑥)
1
2

∙
1

2√𝑥

1

0

𝑑𝑥

∞

𝑛=0

= ∑
(𝑚𝑟)2𝑛

(2𝑛)!
∫ 𝑥𝑛−

1
2 ∙ (1 − 𝑥)−

1
2

1

0

𝑑𝑥

∞

𝑛=0

. 

The integral on the right side of the resulting equation can be calculated using the beta function: 

𝐼(𝑟) = ∑
(𝑚𝑟)2𝑛

(2𝑛)!
∙ 𝐵 (𝑛 +

1

2
;
1

2
)

∞

𝑛=0

= ∑
(𝑚𝑟)2𝑛

(2𝑛)!
∙
Г (𝑛 +

1
2) ∙ Г (

1
2)

Г(𝑛 + 1)

∞

𝑛=0

= 

= ∑ 𝑚2𝑛𝑟2𝑛 ∙
𝜋(2𝑛 − 1)‼

2𝑛 ∙ 𝑛! ∙ (2𝑛)!

∞

𝑛=0

= ∑
𝑚2𝑛𝑟2𝑛𝜋

22𝑛(𝑛!)2

∞

𝑛=0

. 

Therefore,  

𝐼(𝑟) = ∑
𝑚2𝑛𝑟2𝑛𝜋

22𝑛(𝑛!)2

∞

𝑛=0

. 

Problem 2. Evaluate the value of the following integral:  𝐼(𝑏) = ∫ 𝑒−𝑎𝑥2
cos 𝑏𝑥

+∞

0
𝑑𝑥. 



 

American Journal of Interdisciplinary Research and Development 

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38 | P a g e  
 

 

Solution.  

It is known that the given integral depending on the parameter is uniformly convergent 

according to the comparison property. So, the sign of integral can be replaced by the sign of 

sum. To calculate the value of this integral, we use the following Taylor expansion: cos 𝑏𝑥 =

∑
(−1)𝑛(𝑏𝑥)2𝑛

(2𝑛)!
∞
𝑛=0 . In that case, 

𝐼(𝑏) = ∫ 𝑒−𝑎𝑥2
∑

(−1)𝑛(𝑏𝑥)2𝑛

(2𝑛)!

∞

𝑛=0

+∞

0

𝑑𝑥 = ∑ ∫ 𝑒−𝑎𝑥2 (−1)𝑛(𝑏𝑥)2𝑛

(2𝑛)!

+∞

0

𝑑𝑥

∞

𝑛=0

= 

=

[
 
 
 
 
 

𝑎𝑥2 = 𝑡

𝑥 = √
𝑡

𝑎

𝑑𝑥 =
1

2√𝑎𝑡
𝑑𝑡

]
 
 
 
 
 

= ∑
(−1)𝑛𝑏2𝑛

(2𝑛)!
∫ 𝑒−𝑡 (√

𝑡

𝑎
)

2𝑛

(
1

2√𝑎𝑡
)

+∞

0

𝑑𝑡

∞

𝑛=0

= 

= ∑
(−1)𝑛𝑏2𝑛

2(2𝑛)! 𝑎𝑛+
1
2

∫ 𝑒−𝑡𝑡𝑛−
1
2

+∞

0

𝑑𝑡

∞

𝑛=0

= ∑
(−1)𝑛𝑏2𝑛

2(2𝑛)! 𝑎𝑛+
1
2

Г (𝑛 +
1

2
)

∞

𝑛=0

= 

=
1

2
∑

(−1)𝑛𝑏2𝑛

2(2𝑛)! 𝑎𝑛+
1
2

∙
(2𝑛 − 1)‼√𝜋

2𝑛

∞

𝑛=0

=
1

2
√

𝜋

𝛼
∑

(−1)𝑛𝑏2𝑛

2𝑛 ∙ (2𝑛)‼𝑎𝑛

∞

𝑛=0

=
1

2
√

𝜋

𝛼
∑

(−1)𝑛𝑏2𝑛

4𝑛𝑎𝑛 ∙ 𝑛!

∞

𝑛=0

. 

So, the following equality is hold for the given integral: 

𝐼(𝑏) =
1

2
√

𝜋

𝛼
∑ [

(−1)𝑛

𝑛!
∙ (

𝑏2

4𝑎
)

𝑛

]

∞

𝑛=0

. 

It is not difficult to know that the right side of this equation is the Taylor expansion of the 

function 𝑓(𝑏) = 𝑒−
𝑏2

4𝑎. So the value of the given integral is equal to: 

  𝑢(𝑏) = ∫ 𝑒−𝑎𝑥2
cos 𝑏𝑥

+∞

0

𝑑𝑥 =
1

2
√

𝜋

𝛼
∑

(−1)𝑛𝑏2𝑛

4𝑛𝑎𝑛 ∙ 𝑛!

∞

𝑛=0

=
1

2
√

𝜋

𝑎
𝑒−

𝑏2

4𝑎 . 

Problem 3. Evaluate the value of the following integral:   ∫ 𝑓(𝑥)(𝑏 − 𝑥)𝛼𝑏

0
𝑑𝑥.  

Here, 𝛼 > −1 and 𝑓(𝑥) is an analytic function.  

Solution.  

Since 𝛼 > −1 and 𝑓(𝑥) is an analytic function in the given integral, the integral depending on 

this parameter is convergent (Abel's sign). Also, the function 𝑓(𝑥) can be expanded into a 

power series. Thus 

𝑓(𝑥) = ∑
𝑎𝑛𝑥𝑛

𝑛!

∞

𝑛=1

. 

here 𝑎𝑛 − coefficients of extension. Then 



 

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39 | P a g e  
 

 

∫𝑓(𝑥)(𝑏 − 𝑥)𝛼

𝑏

0

𝑑𝑥 = ∫ ∑
𝑎𝑛𝑥𝑛

𝑛!

∞

𝑛=1

(𝑏 − 𝑥)𝛼

𝑏

0

𝑑𝑥 = ∑
𝑎𝑛

𝑛!
∫𝑥𝑛(𝑏 − 𝑥)𝛼

𝑏

0

𝑑𝑥

∞

𝑛=0

= 

  

= [

𝑥 = 𝑏𝑡
𝑑𝑥 = 𝑏𝑑𝑡

𝑥 = 𝑏 → 𝑡 = 1
𝑥 = 0 → 𝑡 = 0

] = ∑
𝑎𝑛

𝑛!
∫𝑏𝑛𝑡𝑛(𝑏 − 𝑏𝑡)𝛼

1

0

𝑏𝑑𝑡

∞

𝑛=1

= ∑
𝑎𝑛𝑏𝑛+𝛼+1

𝑛!
∫ 𝑡𝑛(1 − 𝑡)𝛼

1

0

𝑑𝑡

∞

𝑛=1

= 

   

= ∑
𝑎𝑛𝑏𝑛+𝛼+1

𝑛!
𝐵(𝑛 + 1; 𝛼 + 1)

∞

𝑛=1

= ∑
𝑎𝑛𝑏𝑛+𝛼+1

𝑛!
∙
Г(𝑛 + 1) ∙ Г(𝛼 + 1)

Г(𝑛 + 𝛼 + 2)

∞

𝑛=1

= 

= ∑
𝑎𝑛𝑏𝑛+𝛼+1 ∙ 𝛼Г(𝛼)

(𝑛 + 𝛼 + 2)(𝑛 + 𝛼)Г(𝑛 + 𝛼)

∞

𝑛=1

. 

Problem 4. Evaluate the value of the following integral:   𝐼(𝛼) = ∫ √1 − 𝑥 sin 𝑎𝑥
1

0
𝑑𝑥. 

Solution. To calculate this integral, we use the following Taylor expansion:  

sin 𝛼𝑥 = ∑
(−1)𝑛𝛼2𝑛+1 𝑥2𝑛+1

(2𝑛+1)!
∞
𝑛=1 . Then,    

𝐼(𝛼) = ∫√1 − 𝑥 ∙ sin 𝑎𝑥

1

0

𝑑𝑥 = ∫ ∑
(−1)𝑛𝛼2𝑛+1𝑥2𝑛+1

(2𝑛 + 1)!

∞

𝑛=0

√1 − 𝑥

1

0

𝑑𝑥 = 

= ∑
(−1)𝑛𝛼2𝑛+1

(2𝑛 + 1)!
∫𝑥2𝑛+1(1 − 𝑥)

1
2 

1

0

𝑑𝑥

∞

𝑛=0

= ∑
(−1)𝑛𝛼2𝑛+1

(2𝑛 + 1)!
∙ 𝐵 (2𝑛 + 2,

3

2
)

∞

𝑛=0

= 

∑
(−1)𝑛𝛼2𝑛+1

(2𝑛 + 1)!
∙
Г(2𝑛 + 2) ∙ Г (

3
2)

Г (2𝑛 +
7
2)

∞

𝑛=0

= ∑
(−1)𝑛𝛼2𝑛+1

(2𝑛 + 1)!
∙

(2𝑛 + 1)! ∙
1
2

((2 ∙ (2𝑛 + 2))‼√𝜋)

22𝑛+2

∞

𝑛=0

= 

∑
(−1)𝑛 ∙ 𝛼2𝑛+1 ∙ 22𝑛+2

22𝑛+1 ∙ 𝑛! √𝜋

∞

𝑛=0

=
2

√𝜋
∑

(−1)𝑛 ∙ 𝛼2𝑛+1

𝑛!

∞

𝑛=0

. 

 

Problem 5. Evaluate the value of the following integral:   𝐼(𝛼) = ∫ 𝑥𝛼𝑥1

0
𝑑𝑥. 

Solution. 

To calculate this integral, we first get the following substitution: 

𝐼(𝛼) = ∫𝑥𝛼𝑥

1

0

𝑑𝑥 =

[
 
 
 

𝑥 = 𝑒−𝜉

𝑑𝑥 = −𝑒−𝜉𝑑𝜉
𝑥 → 0   𝜉 → ∞
𝑥 → 1   𝜉 → 0 ]

 
 
 
= ∫ 𝑒−𝜉 ∙ 𝑒−𝜉𝛼𝑒−𝜉

+∞

0

𝑑𝜉. 

 Now we use the following Taylor expansion: 𝑒−𝑥 = ∑
(−1)𝑛𝑥𝑛

𝑛!
∞
𝑛=0 . Then,   



 

American Journal of Interdisciplinary Research and Development 

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Volume 11, Dec., 2022 

40 | P a g e  
 

 

𝐼(𝛼) = ∫ 𝑒−𝜉 ∑
(−1)𝑛(𝛼𝜉𝑒−𝜉)

𝑛

𝑛!

∞

𝑛=0

+∞

0

𝑑𝜉 = ∑
(−1)𝑛𝛼𝑛

𝑛!
∫ 𝜉𝑛𝑒−𝜉(𝑛+1)

+∞

0

𝑑𝜉

∞

𝑛=0

= 

= [
𝜉(𝑛 + 1) = 𝜓

𝑑𝜉 =
𝑑𝜓

𝑛 + 1

] = ∑
(−1)𝑛𝛼𝑛

𝑛!
∫

(
𝜓

𝑛 + 1)
𝑛

𝑒−𝜓

𝑛 + 1

+∞

0

𝑑𝜓

∞

𝑛=0

= 

  

= ∑
(−1)𝑛𝛼𝑛

𝑛! (𝑛 + 1)𝑛+1
∫ 𝜓𝑛𝑒−𝜓

+∞

0

𝑑𝜓 

+∞

𝑛=0

= ∑
(−1)𝑛𝛼𝑛

𝑛! (𝑛 + 1)𝑛+1
Г(𝑛 + 1)

∞

𝑛=0

= ∑
(−1)𝑛−1𝛼𝑛

𝑛𝑛

∞

𝑛=1

. 

 Thus, following equality is holds for given integral: 

𝐼(𝛼) = ∑
(−1)𝑛−1𝛼𝑛

𝑛𝑛

∞

𝑛=1

. 

In particular, ∫ 𝑥𝑥1

0
𝑑𝑥 = ∑

(−1)𝑛−1

𝑛𝑛
∞
𝑛=1  𝑓𝑜𝑟 𝛼 = 1, and ∫

1

𝑥𝑥

1

0
𝑑𝑥 = ∑

1

𝑛𝑛
∞
𝑛=1  𝑓𝑜𝑟 𝛼 = 1. 

Problem 6. Evaluate the value of the following integral:  𝜓(𝛼) = ∫ 𝑒−𝛼𝑥2+𝑥 sin 𝑥
+∞

0
. 

Solution. It is known that the given integral depending on the parameter is uniformly 

convergent for 𝑎 > 0 according to the comparison property. So, the sign of integral can be 

replaced by the sign of sum. To calculate the value of this integral, we use the following Taylor 

expansion: 𝑒𝑥 sin 𝑥 = ∑
2

𝑛
2 sin

𝑛𝜋

4

𝑛!
∙ 𝑥𝑛∞

𝑛=0 . In that case, 

𝜓(𝛼) = ∫ 𝑒−𝛼𝑥2
∑

2
𝑛
2 sin

𝑛𝜋
4

𝑛!
∙ 𝑥𝑛

∞

𝑛=0

+∞

0

𝑑𝑥 = ∑
2

𝑛
2 sin

𝑛𝜋
4

𝑛!
∫ 𝑒−𝛼𝑥2

∙ 𝑥𝑛

+∞

0

𝑑𝑥

+∞

𝑛=0

= 

=

[
 
 
 
 
 
 

𝛼𝑥2 = 𝜉

𝑥 = √
𝜉

𝛼

𝑑𝑥 =
1

2√𝜉𝛼
𝑑𝜉

]
 
 
 
 
 
 

= ∑
2

𝑛
2 sin

𝑛𝜋
4

𝑛!
∫ 𝑒−𝜉 (

𝜉

𝛼
)

𝑛
2

+∞

0

∙
1

2√𝜉𝛼
𝑑𝜉

+∞

𝑛=0

= 

=
1

2
∑

2
𝑛
2 sin

𝑛𝜋
4

𝑛! 𝛼
𝑛+1
2

∫ 𝑒−𝜉

+∞

0

∙ 𝜉
𝑛−1
2 𝑑𝜉

+∞

𝑛=0

=
1

2
∑

2
𝑛
2 sin

𝑛𝜋
4

𝑛! 𝛼
𝑛+1
2

Г (
𝑛 + 1

2
)

+∞

𝑛=0

. 

 

Probem 7. Evaluate the value of the following integral:   𝐼(𝛽) = ∫
𝑒−𝛼𝑥

𝛽+𝑥

+∞

0
𝑑𝑥.  

Solution. It is known that the given integral depending on the parameter is uniformly 

convergent according to the comparison property. So, the sign of integral can be replaced by 



 

American Journal of Interdisciplinary Research and Development 

ISSN Online: 2771-8948 

Website: www.ajird.journalspark.org 

Volume 11, Dec., 2022 

41 | P a g e  
 

 

the sign of sum. To calculate the value of this integral, we use the following Taylor expansion: 
1

1+𝛽𝑥
= ∑ (−1)𝑛𝛽𝑛𝑥𝑛∞

𝑛=0 . In that case, 

𝐼(𝛽) = ∫ 𝑒−𝛼𝑥 ∙ ∑(−1)𝑛𝛽𝑛𝑥𝑛

∞

𝑛=0

+∞

0

𝑑𝑥 = ∑(−1)𝑛𝛽𝑛 ∫ 𝑥𝑛𝑒−𝛼𝑥

+∞

0

𝑑𝑥

∞

𝑛=0

= [

𝛼𝑥 = 𝜙

𝑑𝑥 =
𝑑𝜙

𝛼

] = 

= ∑
(−1)𝑛𝛽𝑛

𝛼𝑛+1
∫ 𝜙𝑛𝑒−𝜙

+∞

0

∞

𝑛=0

𝑑𝜙 = ∑
(−1)𝑛𝛽𝑛

𝛼𝑛+1
Г(𝑛 + 1)

∞

𝑛=0

= ∑
(−1)𝑛𝛽𝑛𝑛!

𝛼𝑛+1

∞

𝑛=0

. 

Thus,  

𝐼(𝛽) = ∑
(−1)𝑛𝛽𝑛𝑛!

𝛼𝑛+1

∞

𝑛=0

. 

Problem 8.  Evaluate the value of the following integral:   ∫ 𝑒−𝛼𝑥𝑝
sin 𝛽𝑥𝑞+∞

0
𝑑𝑥, here 𝑝 > 0,

𝑞 > −1. 

Solution. It is known that the given integral depending on the parameter is uniformly 

convergent for 𝑎 > 0 according to the comparison property. So, the sign of integral can be 

replaced by the sign of sum. To calculate the value of this integral, we use the following Taylor 

expansion: sin 𝑥 = ∑
(−1)𝑛−1𝑥2𝑛−1

(2𝑛−1)!
∞
𝑛=1 . In that case, 

∫ 𝑒−𝛼𝑥𝑝
sin 𝛽𝑥𝑞

+∞

0

𝑑𝑥 = ∫ 𝑒−𝛼𝑥𝑝
∑

(−1)𝑛−1𝛽2𝑛−1𝑥𝑞(2𝑛−1)

(2𝑛 − 1)!

∞

𝑛=1

+∞

0

𝑑𝑥 = 

= ∑
(−1)𝑛−1𝛽2𝑛−1

(2𝑛 − 1)!
∫ 𝑒−𝛼𝑥𝑝

𝑥𝑞(2𝑛−1)

+∞

0

𝑑𝑥

∞

𝑛=1

=

[
 
 
 
 
 
 

𝛼𝑥𝑝 = 𝜉

𝑥 = √
𝜉

𝛼

𝑝

𝑑𝑥 = 𝜉
1
𝑝
−1

∙
1

𝑝 ∙ 𝛼
1
𝑝

𝑑𝜉

]
 
 
 
 
 
 

= 

= ∑
(−1)𝑛−1𝛽(2𝑛−1)

(2𝑛 − 1)! ∙ 𝑝𝛼
𝑞(2𝑛−1)+1

𝑝

∫ 𝑒−𝜉𝜉
𝑞(2𝑛−1)+1

𝑝
−1

+∞

0

𝑑𝑥

∞

𝑛=1

= 

= ∑
(−1)𝑛−1𝛽(2𝑛−1)

(2𝑛 − 1)! 𝑝𝛼
𝑞(2𝑛−1)+1

𝑝

∞

𝑛=1

Г (
𝑞(2𝑛 − 1) + 1

𝑝
). 

In this case, if 𝛼 = 1, 𝛽 = 1, 𝑝 = 2, 𝑞 = 4, we get the following equation: 

∫ 𝑒−𝑥2
sin 𝑥4

+∞

0

𝑑𝑥 = ∑
(−1)𝑛−1𝛽2𝑛−1

(2𝑛 − 1)! 2𝛼
(4(2𝑛−1)+1)

2

Г(
4(2𝑛 − 1) + 1

2
)

∞

𝑛=1

= 



 

American Journal of Interdisciplinary Research and Development 

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Volume 11, Dec., 2022 

42 | P a g e  
 

 

= ∑
(−1)𝑛−1𝛽2𝑛−1

2(2𝑛 − 1)! 𝛼4𝑛−
3
2

Г((4𝑛 − 2) +
1

2
)

∞

𝑛=1

= ∑
(−1)𝑛−1𝛽2𝑛−1

2(2𝑛 − 1)! 𝛼4𝑛−
3
2

∙
(2(4𝑛 − 2) − 1)‼ √𝜋

24𝑛−2

∞

𝑛=1

 

= ∑
(−1)𝑛−1𝛽2𝑛−1

2(2𝑛 − 1)! 𝛼4𝑛−
3
2

∙
(8𝑛 − 3)‼√𝜋

2𝑛

∞

𝑛=1

. 

Problem 9. Evaluate the value of the following integral:   𝐼(𝑝) = ∫
𝑒−𝛼𝑥𝑝

𝑥2+𝑏𝑥+𝑐

∞

0
𝑑𝑥. 

Solution.  It is known that the given integral depending on the parameter is uniformly 

convergent for 𝑎 > 0 according to the comparison property. So, the sign of integral can be 

replaced by the sign of sum. To calculate the value of this integral, we use the following Taylor 

expansion: 
1

𝑥2+𝑏𝑥+𝑐
=

1

√𝑐
∑

𝑥𝑛 sin(𝑛+1)𝜙

sin𝜙
∞
𝑛=0 ,  

here 𝜙 = −arctan√
4𝑐

𝑏2
− 1.  

 ∫
𝑒−𝛼𝑥𝑝

𝑥2 + 𝑏𝑥 + 𝑐

∞

0

𝑑𝑥 = ∫ 𝑒−𝛼𝑥𝑝
∑

𝑥𝑛 sin(𝑛 + 1)𝜙

𝑐 ∙ sin𝜙

∞

𝑛=0

+∞

0

=
1

𝑐
∑

sin(𝑛 + 1)𝜙

sin𝜙 
∫ 𝑒−𝛼𝑥𝑝

𝑥𝑛

+∞

0

𝑑𝑥

∞

𝑛=0

 

=

[
 
 
 
 
 
 
 

𝛼𝑥𝑝 = 𝜉

𝑥 = √(
𝜉

𝛼
)

𝑝

𝑑𝑥 =
𝜉

1
𝑝
−1

𝑝𝛼
1
𝑝

𝑑𝜉

]
 
 
 
 
 
 
 

=
1

𝑐
∑

sin(𝑛 + 1)𝜙

sin𝜙 
∫ 𝑒−𝜉 ∙ (

𝜉

𝛼
)

𝑛
𝑝

∙
𝜉

1
𝑝
−1

𝑝𝛼
1
𝑝

+∞

0

𝑑𝜉

∞

𝑛=0

= 

=
1

𝑐
∑

sin(𝑛 + 1)𝜙

𝑝 ∙ 𝛼
𝑛+1
𝑝 ∙ sin𝜙

∫ 𝑒−𝜉𝜉
𝑛+1
𝑝

−1

+∞

0

∞

𝑛=0

𝑑𝜉 =
1

𝑐
∑

sin(𝑛 + 1)𝜙

𝑝 ∙ 𝛼
𝑛+1
𝑝 ∙ sin 𝜙

Г (
𝑛 + 1

𝑝
)

∞

𝑛=0

. 

 

Problem 10. Evaluate the value of the following integral:   𝐸(𝑘) = ∫ √1 − 𝑘2 sin2 𝜙
𝜋

2
0

𝑑𝜙. 

Solution. To evaluate this integral, we use the following Taylor expansion  

√1 − 𝑥 = 1 − ∑
(2𝑛−3)‼

2𝑛∙𝑛!
𝑥𝑛∞ 

𝑛=1 .  

In that case, 

𝐸(𝑘) = ∫ √1 − 𝑘2 sin2 𝜙

𝜋
2

0

𝑑𝜙 = ∫(1 − ∑
((2𝑛 − 3)‼)

2𝑛𝑛!
∙ (𝑘2 sin2 𝜙)𝑛

∞

𝑛=1

𝜋
2

0

𝑑𝜙 = 

=
𝜋

2
− ∑

(2𝑛 − 3)‼ 𝑘2𝑛

2𝑛𝑛!
∫ sin2𝑛 𝜙

𝜋
2

0

𝑑𝜙

∞

𝑛=1

=

[
 
 
 

∫ sin2𝑛 𝜙

𝜋
2

0

𝑑𝜙 =
(2𝑛 − 1)‼

(2𝑛)‼
∙
𝜋

2

]
 
 
 

= 



 

American Journal of Interdisciplinary Research and Development 

ISSN Online: 2771-8948 

Website: www.ajird.journalspark.org 

Volume 11, Dec., 2022 

43 | P a g e  
 

 

=
𝜋

2
− ∑

(2𝑛 − 3)‼ 𝑘2𝑛

2𝑛𝑛!
∙
(2𝑛 − 1)‼

(2𝑛)‼
∙
𝜋

2

∞

𝑛=1

=
𝜋

2
(1 − ∑ (

(2𝑛 − 1)‼

𝑛!
)

2

∙
𝑘2𝑛

22𝑛(2𝑛 − 1)

∞

𝑛=1

) = 

=
𝜋

2
(1 − ∑ (

(2𝑛 − 1)‼

(2𝑛)!!
)

2

∙
𝑘2𝑛

(2𝑛 − 1)

∞

𝑛=1

). 

If we take 𝑘 =
1

2
 in the given integral, the following equality is holds: 

∫ √1 −
1

4
sin2 𝑥

𝜋
2

0

𝑑𝑥 =
𝜋

2
(1 − ∑ (

(2𝑛 − 1)‼

(2𝑛)!!
)

2

∙
1

22𝑛(2𝑛 − 1)

∞

𝑛=1

). 

 

Problem 11. Evaluate the value of the Bessel function:  𝐼(𝑥) =
1

𝜋
∫ cos(𝑥 ∙ sin𝜙)

𝜋

0
𝑑𝜙. 

Solution. To evaluate this integral, we use the following Taylor expansion  

cos 𝑥 = ∑
((−1)𝑛𝑥2𝑛)

(2𝑛)!
∞
𝑛=0 . 

In that case,  

𝐼(𝑥) =
1

𝜋
∫ cos(𝑥 ∙ sin 𝜙)

𝜋

0

𝑑𝜙 =
1

𝜋
∫ ∑

(−1)𝑛(𝑥 ∙ sin𝜙)2𝑛

(2𝑛)!

∞

𝑛=0

𝜋

0

𝑑𝜙 = 

=
1

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∫ sin2n 𝜙

𝜋

0

𝑑𝜙

∞

𝑛=0

=
1

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∫ sin2n 𝜙

𝜋
2

0

𝑑𝜙

∞

𝑛=0

+
1

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∫ sin2𝑛 𝜙

𝜋

𝜋
2

𝑑𝜙

∞

𝑛=0

. 

In the second integral on the right side of the last equality, we can define it as 𝜙 =
𝜋

2
+ 𝜓. In 

that case, 

𝐼(𝑥) =
1

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∫ sin2n 𝜙

𝜋
2

0

𝑑𝜙

∞

𝑛=0

+
1

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∫ cos2𝑛 𝜓

𝜋
2

0

𝑑𝜓

∞

𝑛=0

= 

=
2

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∫ sin2n 𝜙

𝜋
2

0

𝑑𝜙

∞

𝑛=0

=
2

𝜋
∑

(−1)𝑛𝑥2𝑛

(2𝑛)!
∙
(2𝑛 − 1)‼

(2𝑛)‼
∙
𝜋

2

∞

𝑛=0

= 

= ∑
(−1)𝑛𝑥2𝑛

(2𝑛)!
∙
(2𝑛 − 1)‼

(2𝑛)‼

∞

𝑛=0

= ∑
(−1)𝑛𝑥2𝑛

(2𝑛)!
∙
(2𝑛 − 1)‼

2𝑛 ∙ 𝑛!

∞

𝑛=0

. 

 



 

American Journal of Interdisciplinary Research and Development 

ISSN Online: 2771-8948 

Website: www.ajird.journalspark.org 

Volume 11, Dec., 2022 

44 | P a g e  
 

 

Similar expansions can be successfully used to approximate integrals that cannot be expressed 

in finite form and to compile tables of their values. 

 

References 

1. Alimov Sh., Ashurov R. Matematik tahlil. 3-qism. “Mumtoz so`z”, Toshkent, 2018. 

2. Демидович Б.П. Сборник задач и упражнений по математическому анализу. 

Издательство ЧеРо, 13-е издание. 1997, Москва. 

3. Sa’dullayev A., Mansurov H., Xudoyberganov G. va b.q. Matematik analiz kursidan misol 

va masalalar to`plami. 3-qism. “O`zbekiston” nashriyoti. Toshkent, 1993. 

4. Weisstein, Eric W. "Taylor Series". MathWorld. 

5. Guoning Wu.  Integrals Depending on a Parameter. China University of Petroleum-

Beijing 2017.9. 

 

 

https://en.wikipedia.org/wiki/Eric_W._Weisstein
https://mathworld.wolfram.com/TaylorSeries.html
https://en.wikipedia.org/wiki/MathWorld