Prove $int_0^inftyoperatorname{sech} x,dx=pi/2$, and deduce $int_0^1operatorname{sech}^{-1}x,dx$
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
asked 2 days ago
Abec
145
add a comment |
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
asked 2 days ago
Abec
145
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago
add a comment |
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
asked 2 days ago
Abec
145
Prove $$int_0^inftyoperatorname{sech} x,dx=pi/2$$ and deduce $$int_0^1operatorname{sech}^{-1}x,dx$$
I can prove the first statement (see below), but I was unable to deduce the value of the second integral.
improper-integrals trigonometric-integrals
improper-integrals trigonometric-integrals
asked 2 days ago
Abec
145
asked 2 days ago
Abec
145
edited yesterday
asked 2 days ago
Abec
145
asked 2 days ago
Abec
145
asked 2 days ago
Abec
145
145
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago
add a comment |
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered 2 days ago
Jacky Chong
17.8k21128
add a comment |
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
answered yesterday
DavidG
1,836619
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%2f3059517%2fprove-int-0-infty-operatornamesech-x-dx-pi-2-and-deduce-int-01-opera%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered 2 days ago
Jacky Chong
17.8k21128
add a comment |
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered 2 days ago
Jacky Chong
17.8k21128
add a comment |
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered 2 days ago
Jacky Chong
17.8k21128
Note that by setting $u = operatorname{sech}^{-1} x$ we have that
begin{align}
int^1_0 operatorname{sech}^{-1} x dx = int^infty_0 u tanh u operatorname{sech} u du
end{align}
then by integration by parts you should be done.
answered 2 days ago
Jacky Chong
17.8k21128
edited 2 days ago
answered 2 days ago
Jacky Chong
17.8k21128
answered 2 days ago
Jacky Chong
17.8k21128
answered 2 days ago
Jacky Chong
17.8k21128
17.8k21128
add a comment |
add a comment |
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
answered yesterday
DavidG
1,836619
add a comment |
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
answered yesterday
DavidG
1,836619
add a comment |
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
answered yesterday
DavidG
1,836619
When it comes to the second part of your question:
begin{equation}
I = int f^{-1}(t):dt
end{equation}
Let $x = f^{-1}(t)$ or $t = f(x)$, then
begin{equation}
I = int x cdot f'(x):dx
end{equation}
Integrate by parts:
begin{align}
v'(x) &= f'(x) & u(x) &= x \
v(x) &= f(x) & u'(x) &= 1
end{align}
Thus,
begin{align}
I &= int x cdot f'(x):dx = x cdot f(x) - int f(x) cdot 1 :dx \
&= x cdot f(x) - F(x)
end{align}
Where $F'(x) = f(x)$. As $x = f^{-1}(t)$ we have
begin{align}
I &= int f^{-1}(t):dt = x cdot f(x) - F(x) = t cdot f^{-1}(t) - Fleft(f^{-1}(t)right)
end{align}
So in this case $x = operatorname{arcsech}(t) rightarrow f(x) = t$.
begin{equation}
F'(x) = operatorname{sech}(x) rightarrow F(x) = 2arctanleft(tanhleft(frac{x}{2}right) right)
end{equation}
Hence,
begin{equation}
Fleft(operatorname{arcsech}(x)right) = 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{2}right) right)
end{equation}
And thus:
begin{equation}
int operatorname{arcsech}(t) :dx = tcdot operatorname{arcsech}(t) - 2arctanleft(tanhleft(frac{operatorname{arcsech}(x)}{t}right) right) + C
end{equation}
Where $C$ is the constant of integration.
answered yesterday
DavidG
1,836619
answered yesterday
DavidG
1,836619
answered yesterday
DavidG
1,836619
answered yesterday
DavidG
1,836619
1,836619
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3059517%2fprove-int-0-infty-operatornamesech-x-dx-pi-2-and-deduce-int-01-opera%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
Shouldn't the integral be equal to $frac{pi}{2}$ instead of $pi$? And what's your question, I don't understand what you're asking.
– Matti P.
2 days ago
For the integral of the inverse function, maybe you can use a geometric interpretation! If you graph sech and arcsech, and you know the area under sech ...
– Matti P.
2 days ago
I uploaded the question which is q13. I successfully do the first part but i do no know how to deduce the second part from there. Apparently the arcsech is undefined for 0 to 1 if the i tried to interpret graphically
– Abec
2 days ago