How to show that $int_0^1x^{-x}dx = sum_{n=1}^infty n^{-n}$? [duplicate] The 2019 Stack...
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How to show that $int_0^1x^{-x}dx = sum_{n=1}^infty n^{-n}$? [duplicate]
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series as an integral (sophomore's dream)Sophomore's dream: $displaystyleint_0^{1} x^{-x} ; dx = sum_{n=1}^infty n^{-n}$Numerical Integration: The degree of accuracy of a quadratureIntegrate Using Gauss Laguerre QuadratureHow does one derive Runge Kutta methods from polynomial interpolation?Show that $lim_{nto infty} sum_{k=1}^n frac n{n^2+k^2}=frac pi 4$Prove interval of convergence for the series $sum_{k=1}^infty frac{(k-1)!}{k^k} x^{k-1}$Prove the integral $int_0^1 frac{H_t}{t}dt=sum_{k=1}^{infty} frac{ln (1+frac{1}{k})}{k}$Real Analysis Methodologies to show $gamma =2int_0^infty frac{cos(x^2)-cos(x)}{x},dx$Numerical Analysis Solving SystemsIs $int_0^infty frac{cos(x)}{sqrt{x}}$ lebesgue integrable and improper Riemann integrable?Calculate and provide some justification to get an approximation of $Re(int_0^1(log x)^{-operatorname{li}(x)}dx)$
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This question already has an answer here:
Series as an integral (sophomore's dream)
3 answers
How would I go about showing that $int_0^1x^{-x}dx = sum_{n=1}^infty n^{-n}$
Right now my numerical analysis class is covering gaussian quadrature but we have also covered interpolation. I'm not sure how to prove this equality without using an estimation for the lefthand integral
real-analysis definite-integrals summation numerical-methods
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
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marked as duplicate by Martin R, Yanko, D. Thomine, Community♦ Mar 22 at 21:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Series as an integral (sophomore's dream)
3 answers
How would I go about showing that $int_0^1x^{-x}dx = sum_{n=1}^infty n^{-n}$
Right now my numerical analysis class is covering gaussian quadrature but we have also covered interpolation. I'm not sure how to prove this equality without using an estimation for the lefthand integral
real-analysis definite-integrals summation numerical-methods
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
$endgroup$
marked as duplicate by Martin R, Yanko, D. Thomine, Community♦ Mar 22 at 21:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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math.stackexchange.com/questions/836147/…
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– Martin R
Mar 22 at 21:27
2
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This is known as Sophomore's dream. See this for example: math.stackexchange.com/questions/237513/…. The first link (Wikipedia) also has a proof section.
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– Minus One-Twelfth
Mar 22 at 21:28
add a comment |
$begingroup$
This question already has an answer here:
Series as an integral (sophomore's dream)
3 answers
How would I go about showing that $int_0^1x^{-x}dx = sum_{n=1}^infty n^{-n}$
Right now my numerical analysis class is covering gaussian quadrature but we have also covered interpolation. I'm not sure how to prove this equality without using an estimation for the lefthand integral
real-analysis definite-integrals summation numerical-methods
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
$endgroup$
This question already has an answer here:
Series as an integral (sophomore's dream)
3 answers
How would I go about showing that $int_0^1x^{-x}dx = sum_{n=1}^infty n^{-n}$
Right now my numerical analysis class is covering gaussian quadrature but we have also covered interpolation. I'm not sure how to prove this equality without using an estimation for the lefthand integral
This question already has an answer here:
Series as an integral (sophomore's dream)
3 answers
real-analysis definite-integrals summation numerical-methods
real-analysis definite-integrals summation numerical-methods
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
asked Mar 22 at 21:26
nilay neeranjunnilay neeranjun
11
11
marked as duplicate by Martin R, Yanko, D. Thomine, Community♦ Mar 22 at 21:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Yanko, D. Thomine, Community♦ Mar 22 at 21:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
math.stackexchange.com/questions/836147/…
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– Martin R
Mar 22 at 21:27
2
$begingroup$
This is known as Sophomore's dream. See this for example: math.stackexchange.com/questions/237513/…. The first link (Wikipedia) also has a proof section.
$endgroup$
– Minus One-Twelfth
Mar 22 at 21:28
add a comment |
$begingroup$
math.stackexchange.com/questions/836147/…
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– Martin R
Mar 22 at 21:27
2
$begingroup$
This is known as Sophomore's dream. See this for example: math.stackexchange.com/questions/237513/…. The first link (Wikipedia) also has a proof section.
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– Minus One-Twelfth
Mar 22 at 21:28
$begingroup$
math.stackexchange.com/questions/836147/…
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– Martin R
Mar 22 at 21:27
$begingroup$
math.stackexchange.com/questions/836147/…
$endgroup$
– Martin R
Mar 22 at 21:27
2
2
$begingroup$
This is known as Sophomore's dream. See this for example: math.stackexchange.com/questions/237513/…. The first link (Wikipedia) also has a proof section.
$endgroup$
– Minus One-Twelfth
Mar 22 at 21:28
$begingroup$
This is known as Sophomore's dream. See this for example: math.stackexchange.com/questions/237513/…. The first link (Wikipedia) also has a proof section.
$endgroup$
– Minus One-Twelfth
Mar 22 at 21:28
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1 Answer
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Substitute $y=-ln x$ so the integral becomes $$int_0^1exp(-xln x)dx=sum_{mge 0}frac{(-1)^m}{m!}int_0^1 x^mln^m xdx\=sum_{mge 0}frac{1}{m!}int_0^infty y^mexp[-(m+1)y] dy=sum_{mge 0}frac{1}{(m+1)^{m+1}}.$$
edited Mar 22 at 21:34
clathratus
5,1141439
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Substitute $y=-ln x$ so the integral becomes $$int_0^1exp(-xln x)dx=sum_{mge 0}frac{(-1)^m}{m!}int_0^1 x^mln^m xdx\=sum_{mge 0}frac{1}{m!}int_0^infty y^mexp[-(m+1)y] dy=sum_{mge 0}frac{1}{(m+1)^{m+1}}.$$
edited Mar 22 at 21:34
clathratus
5,1141439
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
$endgroup$
add a comment |
$begingroup$
Substitute $y=-ln x$ so the integral becomes $$int_0^1exp(-xln x)dx=sum_{mge 0}frac{(-1)^m}{m!}int_0^1 x^mln^m xdx\=sum_{mge 0}frac{1}{m!}int_0^infty y^mexp[-(m+1)y] dy=sum_{mge 0}frac{1}{(m+1)^{m+1}}.$$
edited Mar 22 at 21:34
clathratus
5,1141439
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
$endgroup$
add a comment |
$begingroup$
Substitute $y=-ln x$ so the integral becomes $$int_0^1exp(-xln x)dx=sum_{mge 0}frac{(-1)^m}{m!}int_0^1 x^mln^m xdx\=sum_{mge 0}frac{1}{m!}int_0^infty y^mexp[-(m+1)y] dy=sum_{mge 0}frac{1}{(m+1)^{m+1}}.$$
edited Mar 22 at 21:34
clathratus
5,1141439
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
$endgroup$
Substitute $y=-ln x$ so the integral becomes $$int_0^1exp(-xln x)dx=sum_{mge 0}frac{(-1)^m}{m!}int_0^1 x^mln^m xdx\=sum_{mge 0}frac{1}{m!}int_0^infty y^mexp[-(m+1)y] dy=sum_{mge 0}frac{1}{(m+1)^{m+1}}.$$
edited Mar 22 at 21:34
clathratus
5,1141439
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
edited Mar 22 at 21:34
clathratus
5,1141439
edited Mar 22 at 21:34
clathratus
5,1141439
edited Mar 22 at 21:34
clathratus
5,1141439
5,1141439
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
answered Mar 22 at 21:31
J.G.J.G.
33.4k23252
33.4k23252
add a comment |
add a comment |
$begingroup$
math.stackexchange.com/questions/836147/…
$endgroup$
– Martin R
Mar 22 at 21:27
2
$begingroup$
This is known as Sophomore's dream. See this for example: math.stackexchange.com/questions/237513/…. The first link (Wikipedia) also has a proof section.
$endgroup$
– Minus One-Twelfth
Mar 22 at 21:28