How to integrate $-int_{-pi/2}^{pi/2}{frac{sin x}{1+e^x}dx}$?Evaluate the integral $int^{frac{pi}{2}}_0...
Distinction between apt-cache and dpkg -l
How to secure an aircraft at a transient parking space?
Virginia employer terminated employee and wants signing bonus returned
Can you reject a postdoc offer after the PI has paid a large sum for flights/accommodation for your visit?
How to draw cubes in a 3 dimensional plane
Why does the negative sign arise in this thermodynamic relation?
Hotkey (or other quick way) to insert a keyframe for only one component of a vector-valued property?
How can I get players to stop ignoring or overlooking the plot hooks I'm giving them?
What are actual Tesla M60 models used by AWS?
How to detect if C code (which needs 'extern C') is compiled in C++
Is "conspicuously missing" or "conspicuously" the subject of this sentence?
weren't playing vs didn't play
Why would one plane in this picture not have gear down yet?
Accepted offer letter, position changed
Bash script should only kill those instances of another script's that it has launched
Am I not good enough for you?
Word for a person who has no opinion about whether god exists
'The literal of type int is out of range' con número enteros pequeños (2 dígitos)
Are tamper resistant receptacles really safer?
Signed and unsigned numbers
Motivation for Zeta Function of an Algebraic Variety
In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?
PTIJ: wiping amalek’s memory?
PTIJ: Should I kill my computer after installing software?
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 $int_pi^inftyfrac{sin(x^2)}{sqrt{x^2-pi ^2}} , dx$
$begingroup$
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
edited Feb 26 at 14:03
Tom Himler
942314
asked Feb 26 at 12:32
Utkarsh VermaUtkarsh Verma
1147
$endgroup$
|
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
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
edited Feb 26 at 14:03
Tom Himler
942314
asked Feb 26 at 12:32
Utkarsh VermaUtkarsh Verma
1147
$endgroup$
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
definite-integrals
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
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
edited Feb 26 at 14:03
Tom Himler
942314
942314
asked Feb 26 at 12:32
Utkarsh VermaUtkarsh Verma
1147
asked Feb 26 at 12:32
Utkarsh VermaUtkarsh Verma
1147
asked Feb 26 at 12:32
Utkarsh VermaUtkarsh Verma
1147
1147
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
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
1
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
@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
1 Answer
1
active
oldest
votes
$begingroup$
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}
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3127400%2fhow-to-integrate-int-pi-2-pi-2-frac-sin-x1exdx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}
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}
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}
answered 2 days ago
DavidGDavidG
2,5561726
$endgroup$
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}
answered 2 days ago
DavidGDavidG
2,5561726
answered 2 days ago
DavidGDavidG
2,5561726
answered 2 days ago
DavidGDavidG
2,5561726
answered 2 days ago
DavidGDavidG
2,5561726
2,5561726
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3127400%2fhow-to-integrate-int-pi-2-pi-2-frac-sin-x1exdx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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