Mixing
Integral involving $u^{-u}$
$begingroup$
I'm trying to solve the following definite integral
$$int_0^{infty}u^{-u+a},du$$
with $a>0$.
Using integration by parts I've arrived
$$int_0^{infty}u^{-u+a},du=frac{1}{1+a}int_0^{infty}u^{-u+a+1}(1+ln(u)),du$$
which is a worst integral.
Maybe a change of variable, but I don't know how to follow.
definite-integrals
asked Jan 31 at 1:15
popipopi
14512
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the following definite integral
$$int_0^{infty}u^{-u+a},du$$
with $a>0$.
Using integration by parts I've arrived
$$int_0^{infty}u^{-u+a},du=frac{1}{1+a}int_0^{infty}u^{-u+a+1}(1+ln(u)),du$$
which is a worst integral.
Maybe a change of variable, but I don't know how to follow.
definite-integrals
asked Jan 31 at 1:15
popipopi
14512
$endgroup$
1
$begingroup$
I'm afraid there is no elementary integral for $u^{-u}$
$endgroup$
– Bernard Massé
Jan 31 at 1:26
$begingroup$
Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions?
$endgroup$
– popi
Jan 31 at 1:28
$begingroup$
Notice that the minimum value of the integral correspond to $a approx log(2)$. Why ? This is a question.
$endgroup$
– Claude Leibovici
Jan 31 at 6:07
add a comment |
$begingroup$
I'm trying to solve the following definite integral
$$int_0^{infty}u^{-u+a},du$$
with $a>0$.
Using integration by parts I've arrived
$$int_0^{infty}u^{-u+a},du=frac{1}{1+a}int_0^{infty}u^{-u+a+1}(1+ln(u)),du$$
which is a worst integral.
Maybe a change of variable, but I don't know how to follow.
definite-integrals
asked Jan 31 at 1:15
popipopi
14512
$endgroup$
I'm trying to solve the following definite integral
$$int_0^{infty}u^{-u+a},du$$
with $a>0$.
Using integration by parts I've arrived
$$int_0^{infty}u^{-u+a},du=frac{1}{1+a}int_0^{infty}u^{-u+a+1}(1+ln(u)),du$$
which is a worst integral.
Maybe a change of variable, but I don't know how to follow.
definite-integrals
definite-integrals
asked Jan 31 at 1:15
popipopi
14512
asked Jan 31 at 1:15
popipopi
14512
asked Jan 31 at 1:15
popipopi
14512
asked Jan 31 at 1:15
popipopi
14512
asked Jan 31 at 1:15
popipopi
14512
14512
1
$begingroup$
I'm afraid there is no elementary integral for $u^{-u}$
$endgroup$
– Bernard Massé
Jan 31 at 1:26
$begingroup$
Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions?
$endgroup$
– popi
Jan 31 at 1:28
$begingroup$
Notice that the minimum value of the integral correspond to $a approx log(2)$. Why ? This is a question.
$endgroup$
– Claude Leibovici
Jan 31 at 6:07
add a comment |
1
$begingroup$
I'm afraid there is no elementary integral for $u^{-u}$
$endgroup$
– Bernard Massé
Jan 31 at 1:26
$begingroup$
Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions?
$endgroup$
– popi
Jan 31 at 1:28
$begingroup$
Notice that the minimum value of the integral correspond to $a approx log(2)$. Why ? This is a question.
$endgroup$
– Claude Leibovici
Jan 31 at 6:07
1
1
$begingroup$
I'm afraid there is no elementary integral for $u^{-u}$
$endgroup$
– Bernard Massé
Jan 31 at 1:26
$begingroup$
I'm afraid there is no elementary integral for $u^{-u}$
$endgroup$
– Bernard Massé
Jan 31 at 1:26
$begingroup$
Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions?
$endgroup$
– popi
Jan 31 at 1:28
$begingroup$
Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions?
$endgroup$
– popi
Jan 31 at 1:28
$begingroup$
Notice that the minimum value of the integral correspond to $a approx log(2)$. Why ? This is a question.
$endgroup$
– Claude Leibovici
Jan 31 at 6:07
$begingroup$
Notice that the minimum value of the integral correspond to $a approx log(2)$. Why ? This is a question.
$endgroup$
– Claude Leibovici
Jan 31 at 6:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This looks even worse than the sophomore's dream and only numerical methods could be used.
Let $$I_a=int_0^{infty}u^{-u+a},du$$
For small values of $a$, you would get
$$left(
begin{array}{cc}
a & I_a \
0.00 & 1.99546 \
0.25 & 1.81433 \
0.50 & 1.73133 \
0.75 & 1.71513 \
1.00 & 1.75183 \
1.25 & 1.83622 \
1.50 & 1.96842 \
1.75 & 2.15247 \
2.00 & 2.39609
end{array}
right)$$
For large values of $a$, the integral varies extremely fast
$$left(
begin{array}{cc}
a & I_a \
0 & 1.99546 \
1 & 1.75183 \
2 & 2.39609 \
3 & 4.27169 \
4 & 9.22902 \
5 & 23.2062 \
6 & 66.1712 \
7 & 210.120 \
8 & 733.083 \
9 & 2781.12 \
10 & 11378.2
end{array}
right)$$
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
This looks even worse than the sophomore's dream and only numerical methods could be used.
Let $$I_a=int_0^{infty}u^{-u+a},du$$
For small values of $a$, you would get
$$left(
begin{array}{cc}
a & I_a \
0.00 & 1.99546 \
0.25 & 1.81433 \
0.50 & 1.73133 \
0.75 & 1.71513 \
1.00 & 1.75183 \
1.25 & 1.83622 \
1.50 & 1.96842 \
1.75 & 2.15247 \
2.00 & 2.39609
end{array}
right)$$
For large values of $a$, the integral varies extremely fast
$$left(
begin{array}{cc}
a & I_a \
0 & 1.99546 \
1 & 1.75183 \
2 & 2.39609 \
3 & 4.27169 \
4 & 9.22902 \
5 & 23.2062 \
6 & 66.1712 \
7 & 210.120 \
8 & 733.083 \
9 & 2781.12 \
10 & 11378.2
end{array}
right)$$
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
This looks even worse than the sophomore's dream and only numerical methods could be used.
Let $$I_a=int_0^{infty}u^{-u+a},du$$
For small values of $a$, you would get
$$left(
begin{array}{cc}
a & I_a \
0.00 & 1.99546 \
0.25 & 1.81433 \
0.50 & 1.73133 \
0.75 & 1.71513 \
1.00 & 1.75183 \
1.25 & 1.83622 \
1.50 & 1.96842 \
1.75 & 2.15247 \
2.00 & 2.39609
end{array}
right)$$
For large values of $a$, the integral varies extremely fast
$$left(
begin{array}{cc}
a & I_a \
0 & 1.99546 \
1 & 1.75183 \
2 & 2.39609 \
3 & 4.27169 \
4 & 9.22902 \
5 & 23.2062 \
6 & 66.1712 \
7 & 210.120 \
8 & 733.083 \
9 & 2781.12 \
10 & 11378.2
end{array}
right)$$
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
This looks even worse than the sophomore's dream and only numerical methods could be used.
Let $$I_a=int_0^{infty}u^{-u+a},du$$
For small values of $a$, you would get
$$left(
begin{array}{cc}
a & I_a \
0.00 & 1.99546 \
0.25 & 1.81433 \
0.50 & 1.73133 \
0.75 & 1.71513 \
1.00 & 1.75183 \
1.25 & 1.83622 \
1.50 & 1.96842 \
1.75 & 2.15247 \
2.00 & 2.39609
end{array}
right)$$
For large values of $a$, the integral varies extremely fast
$$left(
begin{array}{cc}
a & I_a \
0 & 1.99546 \
1 & 1.75183 \
2 & 2.39609 \
3 & 4.27169 \
4 & 9.22902 \
5 & 23.2062 \
6 & 66.1712 \
7 & 210.120 \
8 & 733.083 \
9 & 2781.12 \
10 & 11378.2
end{array}
right)$$
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
This looks even worse than the sophomore's dream and only numerical methods could be used.
Let $$I_a=int_0^{infty}u^{-u+a},du$$
For small values of $a$, you would get
$$left(
begin{array}{cc}
a & I_a \
0.00 & 1.99546 \
0.25 & 1.81433 \
0.50 & 1.73133 \
0.75 & 1.71513 \
1.00 & 1.75183 \
1.25 & 1.83622 \
1.50 & 1.96842 \
1.75 & 2.15247 \
2.00 & 2.39609
end{array}
right)$$
For large values of $a$, the integral varies extremely fast
$$left(
begin{array}{cc}
a & I_a \
0 & 1.99546 \
1 & 1.75183 \
2 & 2.39609 \
3 & 4.27169 \
4 & 9.22902 \
5 & 23.2062 \
6 & 66.1712 \
7 & 210.120 \
8 & 733.083 \
9 & 2781.12 \
10 & 11378.2
end{array}
right)$$
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 31 at 5:28
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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1
$begingroup$
I'm afraid there is no elementary integral for $u^{-u}$
$endgroup$
– Bernard Massé
Jan 31 at 1:26
$begingroup$
Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions?
$endgroup$
– popi
Jan 31 at 1:28
$begingroup$
Notice that the minimum value of the integral correspond to $a approx log(2)$. Why ? This is a question.
$endgroup$
– Claude Leibovici
Jan 31 at 6:07