Integration of $int_0^inftye^-(x+x^2)dx$ [closed]

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I have the following integral

$$intlimits_0^inftye^-(x+x^2)dx$$

I am wondering if there is an analytical solution for this integral. Any hints or suggestions?

calculus integration exponential-function

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edited Jul 16 at 18:39

David G. Stork

7,6912929

asked Jul 16 at 18:25

optimal control

2501413

closed as off-topic by Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Write $x^2 + x = (x+1/2)^2 - 1/4$ and substitute $y = x+1/2$.
  – Umberto P.
  Jul 16 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are your thoughts about that? What have you tried so far?
  – Taroccoesbrocco
  Jul 16 at 19:20
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

I have the following integral

$$intlimits_0^inftye^-(x+x^2)dx$$

I am wondering if there is an analytical solution for this integral. Any hints or suggestions?

calculus integration exponential-function

share|cite|improve this question

edited Jul 16 at 18:39

David G. Stork

7,6912929

asked Jul 16 at 18:25

optimal control

2501413

closed as off-topic by Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Write $x^2 + x = (x+1/2)^2 - 1/4$ and substitute $y = x+1/2$.
  – Umberto P.
  Jul 16 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are your thoughts about that? What have you tried so far?
  – Taroccoesbrocco
  Jul 16 at 19:20
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I have the following integral

$$intlimits_0^inftye^-(x+x^2)dx$$

I am wondering if there is an analytical solution for this integral. Any hints or suggestions?

calculus integration exponential-function

share|cite|improve this question

edited Jul 16 at 18:39

David G. Stork

7,6912929

asked Jul 16 at 18:25

optimal control

2501413

I have the following integral

$$intlimits_0^inftye^-(x+x^2)dx$$

I am wondering if there is an analytical solution for this integral. Any hints or suggestions?

calculus integration exponential-function

share|cite|improve this question

edited Jul 16 at 18:39

David G. Stork

7,6912929

asked Jul 16 at 18:25

optimal control

2501413

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 18:39

David G. Stork

7,6912929

edited Jul 16 at 18:39

David G. Stork

7,6912929

edited Jul 16 at 18:39

David G. Stork

7,6912929

7,6912929

asked Jul 16 at 18:25

optimal control

2501413

asked Jul 16 at 18:25

optimal control

2501413

asked Jul 16 at 18:25

optimal control

2501413

2501413

closed as off-topic by Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Write $x^2 + x = (x+1/2)^2 - 1/4$ and substitute $y = x+1/2$.
  – Umberto P.
  Jul 16 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are your thoughts about that? What have you tried so far?
  – Taroccoesbrocco
  Jul 16 at 19:20
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Write $x^2 + x = (x+1/2)^2 - 1/4$ and substitute $y = x+1/2$.
  – Umberto P.
  Jul 16 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are your thoughts about that? What have you tried so far?
  – Taroccoesbrocco
  Jul 16 at 19:20
  
  

  
  
  
  

5

5

Write $x^2 + x = (x+1/2)^2 - 1/4$ and substitute $y = x+1/2$.
– Umberto P.
Jul 16 at 18:33

Write $x^2 + x = (x+1/2)^2 - 1/4$ and substitute $y = x+1/2$.
– Umberto P.
Jul 16 at 18:33

What are your thoughts about that? What have you tried so far?
– Taroccoesbrocco
Jul 16 at 19:20

What are your thoughts about that? What have you tried so far?
– Taroccoesbrocco
Jul 16 at 19:20

add a comment | 

3 Answers
3

active

oldest

votes

up vote
5
down vote

Complete the square in the exponent and change variables to find:

$$frac12 sqrt[4]e sqrtpi texterfcleft(frac12right)$$

share|cite|improve this answer

answered Jul 16 at 18:31

David G. Stork

7,6912929

add a comment | 

up vote
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Well, we have:

$$mathcalI_spacetextnleft(textsright):=int_0^inftyexpleft(-left(textscdot x+x^textnright)right)spacetextdxtag1$$

Now, for the $exp$-function we can write:

$$expleft(-left(textscdot x+x^textnright)right)=expleft(-textscdot x-x^textnright)=expleft(-textscdot xright)cdotexpleft(-x^textnright)tag2$$

So, we get:

$$mathcalI_spacetextnleft(textsright)=int_0^inftyexpleft(-textscdot xright)cdotexpleft(-x^textnright)spacetextdxtag3$$

Which is the Laplace transform of $expleft(-x^textnright)$:

$$mathcalI_spacetextnleft(textsright)=mathcalL_xleft[expleft(-x^textnright)right]_left(textsright)tag4$$

Using:

$$expleft(-x^textnright)=sum_textk=0^inftyfracleft(-x^textnright)^textktextk!=sum_textk=0^inftyfracleft(-1right)^textktextk!cdot x^textncdottextktag5$$

We can rewrite equation $left(4right)$:

$$mathcalI_spacetextnleft(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotmathcalL_xleft[x^textncdottextkright]_left(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotfracGammaleft(1+textncdottextkright)texts^1+textncdottextktag6$$

share|cite|improve this answer

edited Jul 16 at 18:35

answered Jul 16 at 18:33

Jan

21.6k31239

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Oh my goodness... you're using a cannon to swat a fly!
  – David G. Stork
  Jul 16 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it is a very nice method!
  – Jan
  Jul 16 at 18:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

$$I=int_0^inftye^-(x+x^2)dx=int_0^inftye^-[(x+0.5)^2-0.25]dx=e^frac14int_0^inftye^-(x+0.5)^2dx$$
$u=x+frac12, du=dx$
$$therefore I=e^frac14int_frac12^inftye^-u^2du=e^frac14left(fracsqrtpi2-int_0^frac12e^-u^2duright)$$
And I think the final integral at the end can maybe expressed in terms of $erf(x)$ or $erfc(x)$

share|cite|improve this answer

answered Jul 16 at 19:43

Henry Lee

49210

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote

Complete the square in the exponent and change variables to find:

$$frac12 sqrt[4]e sqrtpi texterfcleft(frac12right)$$

share|cite|improve this answer

answered Jul 16 at 18:31

David G. Stork

7,6912929

add a comment | 

up vote
5
down vote

Complete the square in the exponent and change variables to find:

$$frac12 sqrt[4]e sqrtpi texterfcleft(frac12right)$$

share|cite|improve this answer

answered Jul 16 at 18:31

David G. Stork

7,6912929

add a comment | 

up vote
5
down vote

up vote
5
down vote

Complete the square in the exponent and change variables to find:

$$frac12 sqrt[4]e sqrtpi texterfcleft(frac12right)$$

share|cite|improve this answer

answered Jul 16 at 18:31

David G. Stork

7,6912929

Complete the square in the exponent and change variables to find:

$$frac12 sqrt[4]e sqrtpi texterfcleft(frac12right)$$

share|cite|improve this answer

answered Jul 16 at 18:31

David G. Stork

7,6912929

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 18:31

David G. Stork

7,6912929

answered Jul 16 at 18:31

David G. Stork

7,6912929

answered Jul 16 at 18:31

David G. Stork

7,6912929

7,6912929

add a comment | 

add a comment | 

up vote
1
down vote

Well, we have:

$$mathcalI_spacetextnleft(textsright):=int_0^inftyexpleft(-left(textscdot x+x^textnright)right)spacetextdxtag1$$

Now, for the $exp$-function we can write:

$$expleft(-left(textscdot x+x^textnright)right)=expleft(-textscdot x-x^textnright)=expleft(-textscdot xright)cdotexpleft(-x^textnright)tag2$$

So, we get:

$$mathcalI_spacetextnleft(textsright)=int_0^inftyexpleft(-textscdot xright)cdotexpleft(-x^textnright)spacetextdxtag3$$

Which is the Laplace transform of $expleft(-x^textnright)$:

$$mathcalI_spacetextnleft(textsright)=mathcalL_xleft[expleft(-x^textnright)right]_left(textsright)tag4$$

Using:

$$expleft(-x^textnright)=sum_textk=0^inftyfracleft(-x^textnright)^textktextk!=sum_textk=0^inftyfracleft(-1right)^textktextk!cdot x^textncdottextktag5$$

We can rewrite equation $left(4right)$:

$$mathcalI_spacetextnleft(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotmathcalL_xleft[x^textncdottextkright]_left(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotfracGammaleft(1+textncdottextkright)texts^1+textncdottextktag6$$

share|cite|improve this answer

edited Jul 16 at 18:35

answered Jul 16 at 18:33

Jan

21.6k31239

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Oh my goodness... you're using a cannon to swat a fly!
  – David G. Stork
  Jul 16 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it is a very nice method!
  – Jan
  Jul 16 at 18:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

Well, we have:

$$mathcalI_spacetextnleft(textsright):=int_0^inftyexpleft(-left(textscdot x+x^textnright)right)spacetextdxtag1$$

Now, for the $exp$-function we can write:

$$expleft(-left(textscdot x+x^textnright)right)=expleft(-textscdot x-x^textnright)=expleft(-textscdot xright)cdotexpleft(-x^textnright)tag2$$

So, we get:

$$mathcalI_spacetextnleft(textsright)=int_0^inftyexpleft(-textscdot xright)cdotexpleft(-x^textnright)spacetextdxtag3$$

Which is the Laplace transform of $expleft(-x^textnright)$:

$$mathcalI_spacetextnleft(textsright)=mathcalL_xleft[expleft(-x^textnright)right]_left(textsright)tag4$$

Using:

$$expleft(-x^textnright)=sum_textk=0^inftyfracleft(-x^textnright)^textktextk!=sum_textk=0^inftyfracleft(-1right)^textktextk!cdot x^textncdottextktag5$$

We can rewrite equation $left(4right)$:

$$mathcalI_spacetextnleft(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotmathcalL_xleft[x^textncdottextkright]_left(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotfracGammaleft(1+textncdottextkright)texts^1+textncdottextktag6$$

share|cite|improve this answer

edited Jul 16 at 18:35

answered Jul 16 at 18:33

Jan

21.6k31239

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Oh my goodness... you're using a cannon to swat a fly!
  – David G. Stork
  Jul 16 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it is a very nice method!
  – Jan
  Jul 16 at 18:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Well, we have:

$$mathcalI_spacetextnleft(textsright):=int_0^inftyexpleft(-left(textscdot x+x^textnright)right)spacetextdxtag1$$

Now, for the $exp$-function we can write:

$$expleft(-left(textscdot x+x^textnright)right)=expleft(-textscdot x-x^textnright)=expleft(-textscdot xright)cdotexpleft(-x^textnright)tag2$$

So, we get:

$$mathcalI_spacetextnleft(textsright)=int_0^inftyexpleft(-textscdot xright)cdotexpleft(-x^textnright)spacetextdxtag3$$

Which is the Laplace transform of $expleft(-x^textnright)$:

$$mathcalI_spacetextnleft(textsright)=mathcalL_xleft[expleft(-x^textnright)right]_left(textsright)tag4$$

Using:

$$expleft(-x^textnright)=sum_textk=0^inftyfracleft(-x^textnright)^textktextk!=sum_textk=0^inftyfracleft(-1right)^textktextk!cdot x^textncdottextktag5$$

We can rewrite equation $left(4right)$:

$$mathcalI_spacetextnleft(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotmathcalL_xleft[x^textncdottextkright]_left(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotfracGammaleft(1+textncdottextkright)texts^1+textncdottextktag6$$

share|cite|improve this answer

edited Jul 16 at 18:35

answered Jul 16 at 18:33

Jan

21.6k31239

Well, we have:

$$mathcalI_spacetextnleft(textsright):=int_0^inftyexpleft(-left(textscdot x+x^textnright)right)spacetextdxtag1$$

Now, for the $exp$-function we can write:

$$expleft(-left(textscdot x+x^textnright)right)=expleft(-textscdot x-x^textnright)=expleft(-textscdot xright)cdotexpleft(-x^textnright)tag2$$

So, we get:

$$mathcalI_spacetextnleft(textsright)=int_0^inftyexpleft(-textscdot xright)cdotexpleft(-x^textnright)spacetextdxtag3$$

Which is the Laplace transform of $expleft(-x^textnright)$:

$$mathcalI_spacetextnleft(textsright)=mathcalL_xleft[expleft(-x^textnright)right]_left(textsright)tag4$$

Using:

$$expleft(-x^textnright)=sum_textk=0^inftyfracleft(-x^textnright)^textktextk!=sum_textk=0^inftyfracleft(-1right)^textktextk!cdot x^textncdottextktag5$$

We can rewrite equation $left(4right)$:

$$mathcalI_spacetextnleft(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotmathcalL_xleft[x^textncdottextkright]_left(textsright)=sum_textk=0^inftyfracleft(-1right)^textktextk!cdotfracGammaleft(1+textncdottextkright)texts^1+textncdottextktag6$$

share|cite|improve this answer

edited Jul 16 at 18:35

answered Jul 16 at 18:33

Jan

21.6k31239

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 18:35

edited Jul 16 at 18:35

edited Jul 16 at 18:35

answered Jul 16 at 18:33

Jan

21.6k31239

answered Jul 16 at 18:33

Jan

21.6k31239

answered Jul 16 at 18:33

Jan

21.6k31239

21.6k31239

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Oh my goodness... you're using a cannon to swat a fly!
  – David G. Stork
  Jul 16 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it is a very nice method!
  – Jan
  Jul 16 at 18:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Oh my goodness... you're using a cannon to swat a fly!
  – David G. Stork
  Jul 16 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it is a very nice method!
  – Jan
  Jul 16 at 18:36
  

  
  
  
  

3

3

Oh my goodness... you're using a cannon to swat a fly!
– David G. Stork
Jul 16 at 18:34

Oh my goodness... you're using a cannon to swat a fly!
– David G. Stork
Jul 16 at 18:34

@DavidG.Stork I think it is a very nice method!
– Jan
Jul 16 at 18:36

@DavidG.Stork I think it is a very nice method!
– Jan
Jul 16 at 18:36

add a comment | 

up vote
1
down vote

$$I=int_0^inftye^-(x+x^2)dx=int_0^inftye^-[(x+0.5)^2-0.25]dx=e^frac14int_0^inftye^-(x+0.5)^2dx$$
$u=x+frac12, du=dx$
$$therefore I=e^frac14int_frac12^inftye^-u^2du=e^frac14left(fracsqrtpi2-int_0^frac12e^-u^2duright)$$
And I think the final integral at the end can maybe expressed in terms of $erf(x)$ or $erfc(x)$

share|cite|improve this answer

answered Jul 16 at 19:43

Henry Lee

49210

add a comment | 

up vote
1
down vote

$$I=int_0^inftye^-(x+x^2)dx=int_0^inftye^-[(x+0.5)^2-0.25]dx=e^frac14int_0^inftye^-(x+0.5)^2dx$$
$u=x+frac12, du=dx$
$$therefore I=e^frac14int_frac12^inftye^-u^2du=e^frac14left(fracsqrtpi2-int_0^frac12e^-u^2duright)$$
And I think the final integral at the end can maybe expressed in terms of $erf(x)$ or $erfc(x)$

share|cite|improve this answer

answered Jul 16 at 19:43

Henry Lee

49210

add a comment | 

up vote
1
down vote

up vote
1
down vote

$$I=int_0^inftye^-(x+x^2)dx=int_0^inftye^-[(x+0.5)^2-0.25]dx=e^frac14int_0^inftye^-(x+0.5)^2dx$$
$u=x+frac12, du=dx$
$$therefore I=e^frac14int_frac12^inftye^-u^2du=e^frac14left(fracsqrtpi2-int_0^frac12e^-u^2duright)$$
And I think the final integral at the end can maybe expressed in terms of $erf(x)$ or $erfc(x)$

share|cite|improve this answer

answered Jul 16 at 19:43

Henry Lee

49210

$$I=int_0^inftye^-(x+x^2)dx=int_0^inftye^-[(x+0.5)^2-0.25]dx=e^frac14int_0^inftye^-(x+0.5)^2dx$$
$u=x+frac12, du=dx$
$$therefore I=e^frac14int_frac12^inftye^-u^2du=e^frac14left(fracsqrtpi2-int_0^frac12e^-u^2duright)$$
And I think the final integral at the end can maybe expressed in terms of $erf(x)$ or $erfc(x)$

share|cite|improve this answer

answered Jul 16 at 19:43

Henry Lee

49210

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 19:43

Henry Lee

49210

answered Jul 16 at 19:43

Henry Lee

49210

answered Jul 16 at 19:43

Henry Lee

49210

49210

add a comment | 

add a comment |