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How to integrate $-int_{-pi/2}^{pi/2}{frac{sin x}{1+e^x}dx}$?

Evaluate the integral $int^{frac{pi}{2}}_0 frac{sin^3x}{sin^3x+cos^3x},mathrm dx$.Derivative of integral of $sin (t^2)$Letting $Ntoinfty$ in an integral containing $sin Nq$Closed form of $displaystyleint_{0}^{pi/4}int_{pi/2}^{pi}frac{(cos x-sin x)^{y-2}}{(cos x+sin x)^{y+2}}, dy, dx$Integral $int_{-infty}^{infty} frac{sin^2x}{x^2} e^{ix}dx$Evaluate $int_{frac{-pi}4}^{frac{pi}4}ln(sin x+cos x)mathrm{d}x$Integrating $int_{-infty}^0e^xsin(x)dx$Evaluate $int_0^inftyfrac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx$how to solve $int_{-infty}^t sin(omega_o(t-t'))(1+tanh(frac{t'}{tau}))dt'$?Evaluate $int_{-1}^1frac{sin(x)}{arcsin(x)}dx$Improper integral $int_pi^inftyfrac{sin(x^2)}{sqrt{x^2-pi ^2}} , dx$

0

1

$begingroup$

I need help with this integral:
$$-int_{-pi/2}^{pi/2}{frac{sin x}{1+e^x}dx}$$

I know this might feel like me asking you to do my homework, but it isn't. I'm simply stuck in this problem and can't figure how to solve it even after investing a couple of hours. I also don't have anyone to ask this problem, therefore I'm asking this question here.

definite-integrals

share|cite|improve this question

edited Feb 26 at 14:03

Tom Himler

942314

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  WA says this here $$frac{1}{2} i left(, _2F_1left(-i,1;1-i;-e^{-pi /2}right)+, _2F_1left(-i,1;1-i;-e^{pi /2}right)-, _2F_1left(i,1;1+i;-e^{-pi /2}right)-, _2F_1left(i,1;1+i;-e^{pi /2}right)right)$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 26 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I bet this isn't homework, but how did you arrive at this integral?
  $endgroup$
  – Zacky
  Feb 26 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the denominator is $$1-e^x$$ use math.stackexchange.com/questions/439851/…
  $endgroup$
  – lab bhattacharjee
  Feb 26 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky It was in a textbook of a friend of mine, meant for a college entrance exam(for IIT-JEE). Though I myself believe this question to be out of context for my level, yet I wanted to ensure if that was the case.
  $endgroup$
  – Utkarsh Verma
  Feb 26 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner I had looked up WA already and had no idea what the answer even meant, that's why I ended up asking this on SE.
  $endgroup$
  – Utkarsh Verma
  Feb 26 at 16:08
  

  
  
  
  

 | 
show 2 more comments

0

1

$begingroup$

I need help with this integral:
$$-int_{-pi/2}^{pi/2}{frac{sin x}{1+e^x}dx}$$

I know this might feel like me asking you to do my homework, but it isn't. I'm simply stuck in this problem and can't figure how to solve it even after investing a couple of hours. I also don't have anyone to ask this problem, therefore I'm asking this question here.

definite-integrals

share|cite|improve this question

edited Feb 26 at 14:03

Tom Himler

942314

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  WA says this here $$frac{1}{2} i left(, _2F_1left(-i,1;1-i;-e^{-pi /2}right)+, _2F_1left(-i,1;1-i;-e^{pi /2}right)-, _2F_1left(i,1;1+i;-e^{-pi /2}right)-, _2F_1left(i,1;1+i;-e^{pi /2}right)right)$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 26 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I bet this isn't homework, but how did you arrive at this integral?
  $endgroup$
  – Zacky
  Feb 26 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the denominator is $$1-e^x$$ use math.stackexchange.com/questions/439851/…
  $endgroup$
  – lab bhattacharjee
  Feb 26 at 15:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zacky It was in a textbook of a friend of mine, meant for a college entrance exam(for IIT-JEE). Though I myself believe this question to be out of context for my level, yet I wanted to ensure if that was the case.
  $endgroup$
  – Utkarsh Verma
  Feb 26 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner I had looked up WA already and had no idea what the answer even meant, that's why I ended up asking this on SE.
  $endgroup$
  – Utkarsh Verma
  Feb 26 at 16:08
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I need help with this integral:
$$-int_{-pi/2}^{pi/2}{frac{sin x}{1+e^x}dx}$$

I know this might feel like me asking you to do my homework, but it isn't. I'm simply stuck in this problem and can't figure how to solve it even after investing a couple of hours. I also don't have anyone to ask this problem, therefore I'm asking this question here.

definite-integrals

share|cite|improve this question

edited Feb 26 at 14:03

Tom Himler

942314

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

$endgroup$

share|cite|improve this question

edited Feb 26 at 14:03

Tom Himler

942314

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

share|cite|improve this question

edited Feb 26 at 14:03

Tom Himler

942314

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

edited Feb 26 at 14:03

Tom Himler

942314

edited Feb 26 at 14:03

Tom Himler

942314

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

asked Feb 26 at 12:32

Utkarsh VermaUtkarsh Verma

1147

1 Answer
1

active

oldest

votes

1

$begingroup$

My initial solution (which I've deleted) was incorrect. Turns out this integral is much more difficult. Here is the work I've done (which is NOT a solution):

begin{equation}
I= int_{-frac{pi}{2}}^{frac{pi}{2}} frac{-sin(x)}{1 + e^x}:dxnonumber
end{equation}

Let us consider a more generalised form:
begin{equation}
J(f(x), a) = int_{-a}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Where $f(x)$ is odd, i.e. $f(x) = -f(-x)$. To address this integral we first split into:
begin{equation}
J(f(x), a) = int_{-a}^{0} frac{-f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Considering the first part, we make the substitution $x mapsto -x$:
begin{equation}
int_{a}^{0} frac{-f(-x)}{1 + e^{-x}}-:dx = int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx nonumber
end{equation}
As $f(x)$ is odd we have $f(-x) = -f(x)$, thus,
begin{equation}
int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx nonumber
end{equation}
Now,
begin{equation}
frac{1}{1 + e^{-x}} = frac{e^x}{e^x + 1}nonumber
end{equation}
And so,
begin{equation}
int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dxnonumber
end{equation}
Returning to $J(f(x),a)$ we have
begin{align}
J(f(x), a) &= int_{-a}^{0} frac{f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^{x}f(x) - f(x)}{e^x + 1}:dx = int_0^a left(frac{e^{x} - 1}{e^x + 1}right)f(x):dx nonumber
end{align}
This is probably not of value for this specific integral.

Returning to $I$ we have
begin{equation}
I = Jleft(sin(x), frac{pi}{2}right) = int_0^frac{pi}{2} left(frac{e^{x} - 1}{e^x + 1}right)sin(x):dx nonumber
end{equation}

share|cite|improve this answer

answered 2 days ago

DavidGDavidG

2,5561726

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When I was solving this problem, I was stuck at this point, the $frac{e^x-1}{e^x+1}$ form. Can't anything be done after that?
  $endgroup$
  – Utkarsh Verma
  2 days ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

My initial solution (which I've deleted) was incorrect. Turns out this integral is much more difficult. Here is the work I've done (which is NOT a solution):

begin{equation}
I= int_{-frac{pi}{2}}^{frac{pi}{2}} frac{-sin(x)}{1 + e^x}:dxnonumber
end{equation}

Let us consider a more generalised form:
begin{equation}
J(f(x), a) = int_{-a}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Where $f(x)$ is odd, i.e. $f(x) = -f(-x)$. To address this integral we first split into:
begin{equation}
J(f(x), a) = int_{-a}^{0} frac{-f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Considering the first part, we make the substitution $x mapsto -x$:
begin{equation}
int_{a}^{0} frac{-f(-x)}{1 + e^{-x}}-:dx = int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx nonumber
end{equation}
As $f(x)$ is odd we have $f(-x) = -f(x)$, thus,
begin{equation}
int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx nonumber
end{equation}
Now,
begin{equation}
frac{1}{1 + e^{-x}} = frac{e^x}{e^x + 1}nonumber
end{equation}
And so,
begin{equation}
int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dxnonumber
end{equation}
Returning to $J(f(x),a)$ we have
begin{align}
J(f(x), a) &= int_{-a}^{0} frac{f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^{x}f(x) - f(x)}{e^x + 1}:dx = int_0^a left(frac{e^{x} - 1}{e^x + 1}right)f(x):dx nonumber
end{align}
This is probably not of value for this specific integral.

Returning to $I$ we have
begin{equation}
I = Jleft(sin(x), frac{pi}{2}right) = int_0^frac{pi}{2} left(frac{e^{x} - 1}{e^x + 1}right)sin(x):dx nonumber
end{equation}

share|cite|improve this answer

answered 2 days ago

DavidGDavidG

2,5561726

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When I was solving this problem, I was stuck at this point, the $frac{e^x-1}{e^x+1}$ form. Can't anything be done after that?
  $endgroup$
  – Utkarsh Verma
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

My initial solution (which I've deleted) was incorrect. Turns out this integral is much more difficult. Here is the work I've done (which is NOT a solution):

begin{equation}
I= int_{-frac{pi}{2}}^{frac{pi}{2}} frac{-sin(x)}{1 + e^x}:dxnonumber
end{equation}

Let us consider a more generalised form:
begin{equation}
J(f(x), a) = int_{-a}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Where $f(x)$ is odd, i.e. $f(x) = -f(-x)$. To address this integral we first split into:
begin{equation}
J(f(x), a) = int_{-a}^{0} frac{-f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Considering the first part, we make the substitution $x mapsto -x$:
begin{equation}
int_{a}^{0} frac{-f(-x)}{1 + e^{-x}}-:dx = int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx nonumber
end{equation}
As $f(x)$ is odd we have $f(-x) = -f(x)$, thus,
begin{equation}
int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx nonumber
end{equation}
Now,
begin{equation}
frac{1}{1 + e^{-x}} = frac{e^x}{e^x + 1}nonumber
end{equation}
And so,
begin{equation}
int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dxnonumber
end{equation}
Returning to $J(f(x),a)$ we have
begin{align}
J(f(x), a) &= int_{-a}^{0} frac{f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^{x}f(x) - f(x)}{e^x + 1}:dx = int_0^a left(frac{e^{x} - 1}{e^x + 1}right)f(x):dx nonumber
end{align}
This is probably not of value for this specific integral.

Returning to $I$ we have
begin{equation}
I = Jleft(sin(x), frac{pi}{2}right) = int_0^frac{pi}{2} left(frac{e^{x} - 1}{e^x + 1}right)sin(x):dx nonumber
end{equation}

share|cite|improve this answer

answered 2 days ago

DavidGDavidG

2,5561726

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When I was solving this problem, I was stuck at this point, the $frac{e^x-1}{e^x+1}$ form. Can't anything be done after that?
  $endgroup$
  – Utkarsh Verma
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

My initial solution (which I've deleted) was incorrect. Turns out this integral is much more difficult. Here is the work I've done (which is NOT a solution):

begin{equation}
I= int_{-frac{pi}{2}}^{frac{pi}{2}} frac{-sin(x)}{1 + e^x}:dxnonumber
end{equation}

Let us consider a more generalised form:
begin{equation}
J(f(x), a) = int_{-a}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Where $f(x)$ is odd, i.e. $f(x) = -f(-x)$. To address this integral we first split into:
begin{equation}
J(f(x), a) = int_{-a}^{0} frac{-f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber
end{equation}
Considering the first part, we make the substitution $x mapsto -x$:
begin{equation}
int_{a}^{0} frac{-f(-x)}{1 + e^{-x}}-:dx = int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx nonumber
end{equation}
As $f(x)$ is odd we have $f(-x) = -f(x)$, thus,
begin{equation}
int_{0}^{a} frac{-f(-x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx nonumber
end{equation}
Now,
begin{equation}
frac{1}{1 + e^{-x}} = frac{e^x}{e^x + 1}nonumber
end{equation}
And so,
begin{equation}
int_{0}^{a} frac{f(x)}{1 + e^{-x}}:dx = int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dxnonumber
end{equation}
Returning to $J(f(x),a)$ we have
begin{align}
J(f(x), a) &= int_{-a}^{0} frac{f(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^xf(x)}{1 + e^{x}}:dx + int_{0}^{a} frac{-f(x)}{1 + e^{x}}:dx nonumber \
&= int_{0}^{a} frac{e^{x}f(x) - f(x)}{e^x + 1}:dx = int_0^a left(frac{e^{x} - 1}{e^x + 1}right)f(x):dx nonumber
end{align}
This is probably not of value for this specific integral.

Returning to $I$ we have
begin{equation}
I = Jleft(sin(x), frac{pi}{2}right) = int_0^frac{pi}{2} left(frac{e^{x} - 1}{e^x + 1}right)sin(x):dx nonumber
end{equation}

share|cite|improve this answer

answered 2 days ago

DavidGDavidG

2,5561726

$endgroup$

share|cite|improve this answer

answered 2 days ago

DavidGDavidG

2,5561726

answered 2 days ago

DavidGDavidG

2,5561726

answered 2 days ago

DavidGDavidG

2,5561726