Integral with $(-1)^t$Double integral with integration by partsEvaluate the integral $intfrac{x^2 + 1}{x^3 +...
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Integral with $(-1)^t$
Double integral with integration by partsEvaluate the integral $intfrac{x^2 + 1}{x^3 + 3x + 1} dx$How it comes that integral of odd function is not even?definite integral of a complex functionFredholm integral?finding the integral with substitutionHelp on solving integral equationCalculate the definite integral with $lim$Exstimation of a Lebesgue integralIntegral of KNN distribution
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can you suggest some exercise online where integral have $(-1)^x$ function like
$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$
real-analysis integration complex-analysis
asked Mar 11 at 10:10
user1828958user1828958
526
$endgroup$
add a comment |
$begingroup$
can you suggest some exercise online where integral have $(-1)^x$ function like
$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$
real-analysis integration complex-analysis
asked Mar 11 at 10:10
user1828958user1828958
526
$endgroup$
$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18
3
$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28
add a comment |
$begingroup$
can you suggest some exercise online where integral have $(-1)^x$ function like
$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$
real-analysis integration complex-analysis
asked Mar 11 at 10:10
user1828958user1828958
526
$endgroup$
can you suggest some exercise online where integral have $(-1)^x$ function like
$int g(x) dx$ where $g(x) = (-1)^x h(x)$ or $g(x) = h((-1)^x,x)$
real-analysis integration complex-analysis
real-analysis integration complex-analysis
asked Mar 11 at 10:10
user1828958user1828958
526
asked Mar 11 at 10:10
user1828958user1828958
526
asked Mar 11 at 10:10
user1828958user1828958
526
asked Mar 11 at 10:10
user1828958user1828958
526
asked Mar 11 at 10:10
user1828958user1828958
526
526
$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18
3
$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28
add a comment |
$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18
3
$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28
$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18
$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18
3
3
$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28
$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...
First, recall that $$e^{ix}=cos(x)+isin(x)$$
Thus, with $x=pi$ we have $$e^{ipi}=-1$$
Take $ln$ on both sides to get
$$ipi=ln(-1)$$
and since $$a^x=e^{xln a}$$
We have that
$$(-1)^x=e^{xln(-1)}=e^{ipi x}=cos(pi x)+isin(pi x)$$
Which is easily integrated.
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...
First, recall that $$e^{ix}=cos(x)+isin(x)$$
Thus, with $x=pi$ we have $$e^{ipi}=-1$$
Take $ln$ on both sides to get
$$ipi=ln(-1)$$
and since $$a^x=e^{xln a}$$
We have that
$$(-1)^x=e^{xln(-1)}=e^{ipi x}=cos(pi x)+isin(pi x)$$
Which is easily integrated.
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
$endgroup$
add a comment |
$begingroup$
I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...
First, recall that $$e^{ix}=cos(x)+isin(x)$$
Thus, with $x=pi$ we have $$e^{ipi}=-1$$
Take $ln$ on both sides to get
$$ipi=ln(-1)$$
and since $$a^x=e^{xln a}$$
We have that
$$(-1)^x=e^{xln(-1)}=e^{ipi x}=cos(pi x)+isin(pi x)$$
Which is easily integrated.
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
$endgroup$
add a comment |
$begingroup$
I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...
First, recall that $$e^{ix}=cos(x)+isin(x)$$
Thus, with $x=pi$ we have $$e^{ipi}=-1$$
Take $ln$ on both sides to get
$$ipi=ln(-1)$$
and since $$a^x=e^{xln a}$$
We have that
$$(-1)^x=e^{xln(-1)}=e^{ipi x}=cos(pi x)+isin(pi x)$$
Which is easily integrated.
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
$endgroup$
I don't really know of any special sources dealing with these sorts of integrals, but I can show you how to integrate it...
First, recall that $$e^{ix}=cos(x)+isin(x)$$
Thus, with $x=pi$ we have $$e^{ipi}=-1$$
Take $ln$ on both sides to get
$$ipi=ln(-1)$$
and since $$a^x=e^{xln a}$$
We have that
$$(-1)^x=e^{xln(-1)}=e^{ipi x}=cos(pi x)+isin(pi x)$$
Which is easily integrated.
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
answered Mar 11 at 19:36
clathratusclathratus
5,0251338
5,0251338
add a comment |
add a comment |
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$begingroup$
What is $(-1)^x$ ? It has no sense if $x$ is not an integer.
$endgroup$
– TheSilverDoe
Mar 11 at 10:18
3
$begingroup$
Or $(-1)^x$ has meaning, but complex values. Then you would probably write $exp(i pi x)$ to be clearer.
$endgroup$
– GEdgar
Mar 11 at 10:28