How can I do this double integral?
Im trying to calculate this integral:
$$int_0^1 int_e^{e^y} frac{1}{ln(x)} dx ~dy$$
Wolfram calculates that it is equal to $1-e$. I have no idea how to do this problem.
calculus
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
asked Nov 27 '18 at 16:14
Pedro Capelas
11
add a comment |
Im trying to calculate this integral:
$$int_0^1 int_e^{e^y} frac{1}{ln(x)} dx ~dy$$
Wolfram calculates that it is equal to $1-e$. I have no idea how to do this problem.
calculus
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
asked Nov 27 '18 at 16:14
Pedro Capelas
11
Hint: $frac{1}{ln x} = x cdotfrac {1/x}{ln x}$.
– TZakrevskiy
Nov 27 '18 at 16:17
Changing the order kills $ln x$.
– Tianlalu
Nov 27 '18 at 16:22
@Tianlalu How does it change? $$int_e^{e^y} int_0^{1} frac{1}{ln(x)} dy ~dx = int_e^{e^y} frac{1}{ln(x)} dx$$
– Pedro Capelas
Nov 27 '18 at 16:24
add a comment |
Im trying to calculate this integral:
$$int_0^1 int_e^{e^y} frac{1}{ln(x)} dx ~dy$$
Wolfram calculates that it is equal to $1-e$. I have no idea how to do this problem.
calculus
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
asked Nov 27 '18 at 16:14
Pedro Capelas
11
Im trying to calculate this integral:
$$int_0^1 int_e^{e^y} frac{1}{ln(x)} dx ~dy$$
Wolfram calculates that it is equal to $1-e$. I have no idea how to do this problem.
calculus
calculus
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
asked Nov 27 '18 at 16:14
Pedro Capelas
11
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
asked Nov 27 '18 at 16:14
Pedro Capelas
11
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
edited Nov 27 '18 at 16:15
Tianlalu
3,09621038
3,09621038
asked Nov 27 '18 at 16:14
Pedro Capelas
11
asked Nov 27 '18 at 16:14
Pedro Capelas
11
asked Nov 27 '18 at 16:14
Pedro Capelas
11
11
Hint: $frac{1}{ln x} = x cdotfrac {1/x}{ln x}$.
– TZakrevskiy
Nov 27 '18 at 16:17
Changing the order kills $ln x$.
– Tianlalu
Nov 27 '18 at 16:22
@Tianlalu How does it change? $$int_e^{e^y} int_0^{1} frac{1}{ln(x)} dy ~dx = int_e^{e^y} frac{1}{ln(x)} dx$$
– Pedro Capelas
Nov 27 '18 at 16:24
add a comment |
Hint: $frac{1}{ln x} = x cdotfrac {1/x}{ln x}$.
– TZakrevskiy
Nov 27 '18 at 16:17
Changing the order kills $ln x$.
– Tianlalu
Nov 27 '18 at 16:22
@Tianlalu How does it change? $$int_e^{e^y} int_0^{1} frac{1}{ln(x)} dy ~dx = int_e^{e^y} frac{1}{ln(x)} dx$$
– Pedro Capelas
Nov 27 '18 at 16:24
Hint: $frac{1}{ln x} = x cdotfrac {1/x}{ln x}$.
– TZakrevskiy
Nov 27 '18 at 16:17
Hint: $frac{1}{ln x} = x cdotfrac {1/x}{ln x}$.
– TZakrevskiy
Nov 27 '18 at 16:17
Changing the order kills $ln x$.
– Tianlalu
Nov 27 '18 at 16:22
Changing the order kills $ln x$.
– Tianlalu
Nov 27 '18 at 16:22
@Tianlalu How does it change? $$int_e^{e^y} int_0^{1} frac{1}{ln(x)} dy ~dx = int_e^{e^y} frac{1}{ln(x)} dx$$
– Pedro Capelas
Nov 27 '18 at 16:24
@Tianlalu How does it change? $$int_e^{e^y} int_0^{1} frac{1}{ln(x)} dy ~dx = int_e^{e^y} frac{1}{ln(x)} dx$$
– Pedro Capelas
Nov 27 '18 at 16:24
add a comment |
1 Answer
1
active
oldest
votes
Hint. Note that
$$int_{y=0}^1 left(int_{x=e}^{e^y} frac{1}{ln(x)} dxright)dy=int_{x=e}^1 frac{1}{ln(x)}left(int_{y=?}^{?} dyright)dx.$$
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint. Note that
$$int_{y=0}^1 left(int_{x=e}^{e^y} frac{1}{ln(x)} dxright)dy=int_{x=e}^1 frac{1}{ln(x)}left(int_{y=?}^{?} dyright)dx.$$
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
add a comment |
Hint. Note that
$$int_{y=0}^1 left(int_{x=e}^{e^y} frac{1}{ln(x)} dxright)dy=int_{x=e}^1 frac{1}{ln(x)}left(int_{y=?}^{?} dyright)dx.$$
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
add a comment |
Hint. Note that
$$int_{y=0}^1 left(int_{x=e}^{e^y} frac{1}{ln(x)} dxright)dy=int_{x=e}^1 frac{1}{ln(x)}left(int_{y=?}^{?} dyright)dx.$$
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
Hint. Note that
$$int_{y=0}^1 left(int_{x=e}^{e^y} frac{1}{ln(x)} dxright)dy=int_{x=e}^1 frac{1}{ln(x)}left(int_{y=?}^{?} dyright)dx.$$
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
answered Nov 27 '18 at 16:24
Robert Z
93.3k1061132
93.3k1061132
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
add a comment |
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
That's exactly my doubt, I don't know the limits of this new integral.
– Pedro Capelas
Nov 27 '18 at 16:30
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
Make a drawing of the domain of integration and consider the bounding curves such as $x=e^y$.
– Robert Z
Nov 27 '18 at 16:32
add a comment |
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Hint: $frac{1}{ln x} = x cdotfrac {1/x}{ln x}$.
– TZakrevskiy
Nov 27 '18 at 16:17
Changing the order kills $ln x$.
– Tianlalu
Nov 27 '18 at 16:22
@Tianlalu How does it change? $$int_e^{e^y} int_0^{1} frac{1}{ln(x)} dy ~dx = int_e^{e^y} frac{1}{ln(x)} dx$$
– Pedro Capelas
Nov 27 '18 at 16:24