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An integral I am stuck with?

integral from 0 to 0.25 of arcsin(sqrt(x))?Stuck on Indefinite IntegralA closed-form of $frac{1}{2}int_0^inftyleft[frac{x^2cos x}{cosh 2x-cos x}-frac{2x^2}{e^{4x}-2e^{2x}cos x+1}right],dx$Real analytic methods for the following integralConvergence of $int_0^infty frac{sqrt{x}sin x}{(e^x-1)log(1+x)}dx$Troubles with double integralSeemingly impossible double integral reductionConvergence of an improper integral $I=int_0^infty frac{sin{x}-xcos{x}}{x^alpha}dx$Functional equation with integralStuck on limit of line integral

3

2

$begingroup$

How to work out the integral $$int_0^{2π} e^{e^{ix}} , dx.$$ I am now stuck with this for 2 days please help ! Any hint would be appreciated!

calculus

share|cite|improve this question

edited 30 mins ago

Aditya Garg

asked 53 mins ago

Aditya GargAditya Garg

1358

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried in the past two days, and what is your mathematical background?
  $endgroup$
  – Strants
  49 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure I understand. Do you mean you've transformed the integrand to that form and then can't continue? Or that you are trying to get an integral with that integrand?
  $endgroup$
  – Strants
  41 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I transformed the integrand to the above mentioned formed but couldn't continue
  $endgroup$
  – Aditya Garg
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Was this an exercise someone posed for you, or is it just a problem you made up yourself? Not every function has a closed-form antiderivative, and I suspect that this one doesn't. (It can be solved via contour integration, but that's a much more advanced technique than is taught in high schools.)
  $endgroup$
  – Michael Seifert
  38 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you do that transformation? Where did the "$i$" go?
  $endgroup$
  – Robert Lewis
  38 mins ago
  

  
  
  
  

 | 
show 1 more comment

3

2

$begingroup$

How to work out the integral $$int_0^{2π} e^{e^{ix}} , dx.$$ I am now stuck with this for 2 days please help ! Any hint would be appreciated!

calculus

share|cite|improve this question

edited 30 mins ago

Aditya Garg

asked 53 mins ago

Aditya GargAditya Garg

1358

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried in the past two days, and what is your mathematical background?
  $endgroup$
  – Strants
  49 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure I understand. Do you mean you've transformed the integrand to that form and then can't continue? Or that you are trying to get an integral with that integrand?
  $endgroup$
  – Strants
  41 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I transformed the integrand to the above mentioned formed but couldn't continue
  $endgroup$
  – Aditya Garg
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Was this an exercise someone posed for you, or is it just a problem you made up yourself? Not every function has a closed-form antiderivative, and I suspect that this one doesn't. (It can be solved via contour integration, but that's a much more advanced technique than is taught in high schools.)
  $endgroup$
  – Michael Seifert
  38 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you do that transformation? Where did the "$i$" go?
  $endgroup$
  – Robert Lewis
  38 mins ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

How to work out the integral $$int_0^{2π} e^{e^{ix}} , dx.$$ I am now stuck with this for 2 days please help ! Any hint would be appreciated!

calculus

share|cite|improve this question

edited 30 mins ago

Aditya Garg

asked 53 mins ago

Aditya GargAditya Garg

1358

$endgroup$

share|cite|improve this question

edited 30 mins ago

Aditya Garg

asked 53 mins ago

Aditya GargAditya Garg

1358

share|cite|improve this question

edited 30 mins ago

Aditya Garg

asked 53 mins ago

Aditya GargAditya Garg

1358

asked 53 mins ago

Aditya GargAditya Garg

1358

asked 53 mins ago

Aditya GargAditya Garg

1358

3 Answers
3

active

oldest

votes

7

$begingroup$

One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.

Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=int_0^{2π} e^{large tf} dxRightarrow I'(t)=int_0^{2pi} f e^{large tf}dx$$
But since $(tf)' = left(te^{ix}right)'=it e^{ix} =it f$

$$Rightarrow I'(t)=frac{1}{it} int_0^{2pi} left(e^{tf}right)'dx =frac{e^{tf}}{it}bigg|_0^{2pi} $$
Now $e^{ix}$ is periodic with $T=2pi $ and $e^{2pi i} = e^0 =1$ so we have:$$I'(t)=frac{e^t -e^t}{it}=0$$
And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer.
$$I(t)=I(0)=int_0^{2pi} dx=2pi$$

share|cite|improve this answer

edited 13 mins ago

J. W. Tanner

2,1441117

answered 28 mins ago

ZackyZacky

6,7701958

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks it was a good way of doing that but could you also try to proceed using my transformed integrand
  $endgroup$
  – Aditya Garg
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I couldnt find the post that I remembered but here's a similar one: artofproblemsolving.com/community/…
  $endgroup$
  – Zacky
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AdityaGarg I will try to read it.
  $endgroup$
  – Zacky
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is the brilliant idea! You basically showed first that the Imaginary part of $int_0^{2pi} e^{e^{ix}} dx=0$ with that $2pi-x =x$ substitution and you were left with the real part. I am not sure though if arriving at $int_0^pi cosh(cos x)cos(sin x) dx$ does simplify things (as a combination of trigonometric function and a hyperbolic one is not so nice), but hey it's nice to know that it equals to $pi$./// Afterwards you can proceed like in the link I gave on the second link.
  $endgroup$
  – Zacky
  14 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot I'll try to think and proceed
  $endgroup$
  – Aditya Garg
  12 mins ago
  

  
  
  
  

 | 
show 2 more comments

5

$begingroup$

Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes

$$ int_{C} frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $frac{1}{i}$ and using the resiude theorem then the integral is $2pi i ; Res_0(f) = 2pi i frac{1}{i} = 2pi$

share|cite|improve this answer

answered 32 mins ago

JoseSquareJoseSquare

72012

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmmm seems a bit advance for a person of my skill set
  $endgroup$
  – Aditya Garg
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I didn´t knew it, but I hope in the future will help you!!
  $endgroup$
  – JoseSquare
  25 mins ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

Hint: Use Euler's formula

$$e^{ix}=cos{x}+isin{x}implies e^{e^{ix}}=e^{cos{x}+isin{x}}=e^{cos{x}}e^{isin{x}}$$

Slightly better hint:

You'll have to use the exponential integral to make any sense of this from this point forward.

share|cite|improve this answer

edited 31 mins ago

answered 44 mins ago

R. BurtonR. Burton

575110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was a obvious step a more deeper hint would be appreciated. Although thanks for help
  $endgroup$
  – Aditya Garg
  41 mins ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Quite frankly, I can't even see how this helps. Where to go from here?
  $endgroup$
  – Robert Lewis
  38 mins ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.

Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=int_0^{2π} e^{large tf} dxRightarrow I'(t)=int_0^{2pi} f e^{large tf}dx$$
But since $(tf)' = left(te^{ix}right)'=it e^{ix} =it f$

$$Rightarrow I'(t)=frac{1}{it} int_0^{2pi} left(e^{tf}right)'dx =frac{e^{tf}}{it}bigg|_0^{2pi} $$
Now $e^{ix}$ is periodic with $T=2pi $ and $e^{2pi i} = e^0 =1$ so we have:$$I'(t)=frac{e^t -e^t}{it}=0$$
And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer.
$$I(t)=I(0)=int_0^{2pi} dx=2pi$$

share|cite|improve this answer

edited 13 mins ago

J. W. Tanner

2,1441117

answered 28 mins ago

ZackyZacky

6,7701958

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks it was a good way of doing that but could you also try to proceed using my transformed integrand
  $endgroup$
  – Aditya Garg
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I couldnt find the post that I remembered but here's a similar one: artofproblemsolving.com/community/…
  $endgroup$
  – Zacky
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AdityaGarg I will try to read it.
  $endgroup$
  – Zacky
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is the brilliant idea! You basically showed first that the Imaginary part of $int_0^{2pi} e^{e^{ix}} dx=0$ with that $2pi-x =x$ substitution and you were left with the real part. I am not sure though if arriving at $int_0^pi cosh(cos x)cos(sin x) dx$ does simplify things (as a combination of trigonometric function and a hyperbolic one is not so nice), but hey it's nice to know that it equals to $pi$./// Afterwards you can proceed like in the link I gave on the second link.
  $endgroup$
  – Zacky
  14 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot I'll try to think and proceed
  $endgroup$
  – Aditya Garg
  12 mins ago
  

  
  
  
  

 | 
show 2 more comments

7

$begingroup$

One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.

Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=int_0^{2π} e^{large tf} dxRightarrow I'(t)=int_0^{2pi} f e^{large tf}dx$$
But since $(tf)' = left(te^{ix}right)'=it e^{ix} =it f$

$$Rightarrow I'(t)=frac{1}{it} int_0^{2pi} left(e^{tf}right)'dx =frac{e^{tf}}{it}bigg|_0^{2pi} $$
Now $e^{ix}$ is periodic with $T=2pi $ and $e^{2pi i} = e^0 =1$ so we have:$$I'(t)=frac{e^t -e^t}{it}=0$$
And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer.
$$I(t)=I(0)=int_0^{2pi} dx=2pi$$

share|cite|improve this answer

edited 13 mins ago

J. W. Tanner

2,1441117

answered 28 mins ago

ZackyZacky

6,7701958

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks it was a good way of doing that but could you also try to proceed using my transformed integrand
  $endgroup$
  – Aditya Garg
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I couldnt find the post that I remembered but here's a similar one: artofproblemsolving.com/community/…
  $endgroup$
  – Zacky
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AdityaGarg I will try to read it.
  $endgroup$
  – Zacky
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is the brilliant idea! You basically showed first that the Imaginary part of $int_0^{2pi} e^{e^{ix}} dx=0$ with that $2pi-x =x$ substitution and you were left with the real part. I am not sure though if arriving at $int_0^pi cosh(cos x)cos(sin x) dx$ does simplify things (as a combination of trigonometric function and a hyperbolic one is not so nice), but hey it's nice to know that it equals to $pi$./// Afterwards you can proceed like in the link I gave on the second link.
  $endgroup$
  – Zacky
  14 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot I'll try to think and proceed
  $endgroup$
  – Aditya Garg
  12 mins ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.

Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=int_0^{2π} e^{large tf} dxRightarrow I'(t)=int_0^{2pi} f e^{large tf}dx$$
But since $(tf)' = left(te^{ix}right)'=it e^{ix} =it f$

$$Rightarrow I'(t)=frac{1}{it} int_0^{2pi} left(e^{tf}right)'dx =frac{e^{tf}}{it}bigg|_0^{2pi} $$
Now $e^{ix}$ is periodic with $T=2pi $ and $e^{2pi i} = e^0 =1$ so we have:$$I'(t)=frac{e^t -e^t}{it}=0$$
And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer.
$$I(t)=I(0)=int_0^{2pi} dx=2pi$$

share|cite|improve this answer

edited 13 mins ago

J. W. Tanner

2,1441117

answered 28 mins ago

ZackyZacky

6,7701958

$endgroup$

share|cite|improve this answer

edited 13 mins ago

J. W. Tanner

2,1441117

answered 28 mins ago

ZackyZacky

6,7701958

edited 13 mins ago

J. W. Tanner

2,1441117

edited 13 mins ago

J. W. Tanner

2,1441117

answered 28 mins ago

ZackyZacky

6,7701958

answered 28 mins ago

ZackyZacky

6,7701958

5

$begingroup$

Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes

$$ int_{C} frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $frac{1}{i}$ and using the resiude theorem then the integral is $2pi i ; Res_0(f) = 2pi i frac{1}{i} = 2pi$

share|cite|improve this answer

answered 32 mins ago

JoseSquareJoseSquare

72012

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmmm seems a bit advance for a person of my skill set
  $endgroup$
  – Aditya Garg
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I didn´t knew it, but I hope in the future will help you!!
  $endgroup$
  – JoseSquare
  25 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes

$$ int_{C} frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $frac{1}{i}$ and using the resiude theorem then the integral is $2pi i ; Res_0(f) = 2pi i frac{1}{i} = 2pi$

share|cite|improve this answer

answered 32 mins ago

JoseSquareJoseSquare

72012

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmmm seems a bit advance for a person of my skill set
  $endgroup$
  – Aditya Garg
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I didn´t knew it, but I hope in the future will help you!!
  $endgroup$
  – JoseSquare
  25 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes

$$ int_{C} frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $frac{1}{i}$ and using the resiude theorem then the integral is $2pi i ; Res_0(f) = 2pi i frac{1}{i} = 2pi$

share|cite|improve this answer

answered 32 mins ago

JoseSquareJoseSquare

72012

$endgroup$

share|cite|improve this answer

answered 32 mins ago

JoseSquareJoseSquare

72012

answered 32 mins ago

JoseSquareJoseSquare

72012

answered 32 mins ago

JoseSquareJoseSquare

72012

-1

$begingroup$

Hint: Use Euler's formula

$$e^{ix}=cos{x}+isin{x}implies e^{e^{ix}}=e^{cos{x}+isin{x}}=e^{cos{x}}e^{isin{x}}$$

Slightly better hint:

You'll have to use the exponential integral to make any sense of this from this point forward.

share|cite|improve this answer

edited 31 mins ago

answered 44 mins ago

R. BurtonR. Burton

575110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was a obvious step a more deeper hint would be appreciated. Although thanks for help
  $endgroup$
  – Aditya Garg
  41 mins ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Quite frankly, I can't even see how this helps. Where to go from here?
  $endgroup$
  – Robert Lewis
  38 mins ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

Hint: Use Euler's formula

$$e^{ix}=cos{x}+isin{x}implies e^{e^{ix}}=e^{cos{x}+isin{x}}=e^{cos{x}}e^{isin{x}}$$

Slightly better hint:

You'll have to use the exponential integral to make any sense of this from this point forward.

share|cite|improve this answer

edited 31 mins ago

answered 44 mins ago

R. BurtonR. Burton

575110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was a obvious step a more deeper hint would be appreciated. Although thanks for help
  $endgroup$
  – Aditya Garg
  41 mins ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Quite frankly, I can't even see how this helps. Where to go from here?
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  – Robert Lewis
  38 mins ago
  

  
  
  
  

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Hint: Use Euler's formula

$$e^{ix}=cos{x}+isin{x}implies e^{e^{ix}}=e^{cos{x}+isin{x}}=e^{cos{x}}e^{isin{x}}$$

Slightly better hint:

You'll have to use the exponential integral to make any sense of this from this point forward.

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edited 31 mins ago

answered 44 mins ago

R. BurtonR. Burton

575110

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share|cite|improve this answer

edited 31 mins ago

answered 44 mins ago

R. BurtonR. Burton

575110

answered 44 mins ago

R. BurtonR. Burton

575110

answered 44 mins ago

R. BurtonR. Burton

575110