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What is $ displaystyleintdfracx^41+ e^x dx $?
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Here's the integral I have,
$$ displaystyleintdfracx^41+ e^x dx $$
I tried the usual methods I know, but I failed miserably.
How would you all approach this problem?
calculus integration indefinite-integrals
edited Jul 20 at 21:13
Abcd
2,3761624
asked Jul 20 at 19:55
William
801214
 |Â
show 5 more comments
up vote
4
down vote
favorite
Here's the integral I have,
$$ displaystyleintdfracx^41+ e^x dx $$
I tried the usual methods I know, but I failed miserably.
How would you all approach this problem?
calculus integration indefinite-integrals
edited Jul 20 at 21:13
Abcd
2,3761624
asked Jul 20 at 19:55
William
801214
3
Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral.
– md2perpe
Jul 20 at 19:58
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 20 at 20:00
@md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie)
– William
Jul 20 at 20:01
The integral is very similar to the one here.
– md2perpe
Jul 20 at 20:01
@md2perpe Yea I know, I did check that a while back, but it wasn't much help :(
– William
Jul 20 at 20:03
 |Â
show 5 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Here's the integral I have,
$$ displaystyleintdfracx^41+ e^x dx $$
I tried the usual methods I know, but I failed miserably.
How would you all approach this problem?
calculus integration indefinite-integrals
edited Jul 20 at 21:13
Abcd
2,3761624
asked Jul 20 at 19:55
William
801214
Here's the integral I have,
$$ displaystyleintdfracx^41+ e^x dx $$
I tried the usual methods I know, but I failed miserably.
How would you all approach this problem?
calculus integration indefinite-integrals
edited Jul 20 at 21:13
Abcd
2,3761624
asked Jul 20 at 19:55
William
801214
edited Jul 20 at 21:13
Abcd
2,3761624
edited Jul 20 at 21:13
Abcd
2,3761624
edited Jul 20 at 21:13
Abcd
2,3761624
2,3761624
asked Jul 20 at 19:55
William
801214
asked Jul 20 at 19:55
William
801214
asked Jul 20 at 19:55
William
801214
801214
3
Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral.
– md2perpe
Jul 20 at 19:58
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 20 at 20:00
@md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie)
– William
Jul 20 at 20:01
The integral is very similar to the one here.
– md2perpe
Jul 20 at 20:01
@md2perpe Yea I know, I did check that a while back, but it wasn't much help :(
– William
Jul 20 at 20:03
 |Â
show 5 more comments
3
Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral.
– md2perpe
Jul 20 at 19:58
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 20 at 20:00
@md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie)
– William
Jul 20 at 20:01
The integral is very similar to the one here.
– md2perpe
Jul 20 at 20:01
@md2perpe Yea I know, I did check that a while back, but it wasn't much help :(
– William
Jul 20 at 20:03
3
3
Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral.
– md2perpe
Jul 20 at 19:58
Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral.
– md2perpe
Jul 20 at 19:58
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 20 at 20:00
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 20 at 20:00
@md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie)
– William
Jul 20 at 20:01
@md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie)
– William
Jul 20 at 20:01
The integral is very similar to the one here.
– md2perpe
Jul 20 at 20:01
The integral is very similar to the one here.
– md2perpe
Jul 20 at 20:01
@md2perpe Yea I know, I did check that a while back, but it wasn't much help :(
– William
Jul 20 at 20:03
@md2perpe Yea I know, I did check that a while back, but it wasn't much help :(
– William
Jul 20 at 20:03
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
According to WolframAlpha, you would have to use nonelementary functions such as the polylogarithm function, $operatornameLi_n(x)$:
$$fracx^55 - x^4 log(1 + e^x) - 4 x^3 operatornameLi_2(-e^x) + 12 x^2 operatornameLi_3(-e^x) - 24 x operatornameLi_4(-e^x) + 24 operatornameLi_5(-e^x)$$
edited Jul 20 at 20:16
J.G.
13.2k11424
answered Jul 20 at 20:08
RayDansh
884214
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
add a comment |Â
up vote
4
down vote
$intfrac x^41+e^x dx\
intfrac x^4e^-x1+e^-x dx$
Converting to a geometric series: $frac y1+y = sum_limitsn=1^infty (-y)^n$
$int x^4sum_limitsn=1^infty (-1)^ne^-nx dx$
considering just one term $int x^4e^-nx = (frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
$sum_limitsn=1^infty(-1)^n(frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
It is worth noting
$frac1Gamma(s) int_0^infty frac x^s-1e^x -1 dx = zeta(s)$
answered Jul 20 at 20:41
Doug M
39.2k31749
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
According to WolframAlpha, you would have to use nonelementary functions such as the polylogarithm function, $operatornameLi_n(x)$:
$$fracx^55 - x^4 log(1 + e^x) - 4 x^3 operatornameLi_2(-e^x) + 12 x^2 operatornameLi_3(-e^x) - 24 x operatornameLi_4(-e^x) + 24 operatornameLi_5(-e^x)$$
edited Jul 20 at 20:16
J.G.
13.2k11424
answered Jul 20 at 20:08
RayDansh
884214
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
add a comment |Â
up vote
4
down vote
According to WolframAlpha, you would have to use nonelementary functions such as the polylogarithm function, $operatornameLi_n(x)$:
$$fracx^55 - x^4 log(1 + e^x) - 4 x^3 operatornameLi_2(-e^x) + 12 x^2 operatornameLi_3(-e^x) - 24 x operatornameLi_4(-e^x) + 24 operatornameLi_5(-e^x)$$
edited Jul 20 at 20:16
J.G.
13.2k11424
answered Jul 20 at 20:08
RayDansh
884214
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
add a comment |Â
up vote
4
down vote
up vote
4
down vote
According to WolframAlpha, you would have to use nonelementary functions such as the polylogarithm function, $operatornameLi_n(x)$:
$$fracx^55 - x^4 log(1 + e^x) - 4 x^3 operatornameLi_2(-e^x) + 12 x^2 operatornameLi_3(-e^x) - 24 x operatornameLi_4(-e^x) + 24 operatornameLi_5(-e^x)$$
edited Jul 20 at 20:16
J.G.
13.2k11424
answered Jul 20 at 20:08
RayDansh
884214
According to WolframAlpha, you would have to use nonelementary functions such as the polylogarithm function, $operatornameLi_n(x)$:
$$fracx^55 - x^4 log(1 + e^x) - 4 x^3 operatornameLi_2(-e^x) + 12 x^2 operatornameLi_3(-e^x) - 24 x operatornameLi_4(-e^x) + 24 operatornameLi_5(-e^x)$$
edited Jul 20 at 20:16
J.G.
13.2k11424
answered Jul 20 at 20:08
RayDansh
884214
edited Jul 20 at 20:16
J.G.
13.2k11424
edited Jul 20 at 20:16
J.G.
13.2k11424
edited Jul 20 at 20:16
J.G.
13.2k11424
13.2k11424
answered Jul 20 at 20:08
RayDansh
884214
answered Jul 20 at 20:08
RayDansh
884214
answered Jul 20 at 20:08
RayDansh
884214
884214
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
add a comment |Â
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
+1 .... Is it just me or does anyone here shy away from polylogarithms too?
– William
Jul 20 at 20:22
add a comment |Â
up vote
4
down vote
$intfrac x^41+e^x dx\
intfrac x^4e^-x1+e^-x dx$
Converting to a geometric series: $frac y1+y = sum_limitsn=1^infty (-y)^n$
$int x^4sum_limitsn=1^infty (-1)^ne^-nx dx$
considering just one term $int x^4e^-nx = (frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
$sum_limitsn=1^infty(-1)^n(frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
It is worth noting
$frac1Gamma(s) int_0^infty frac x^s-1e^x -1 dx = zeta(s)$
answered Jul 20 at 20:41
Doug M
39.2k31749
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
add a comment |Â
up vote
4
down vote
$intfrac x^41+e^x dx\
intfrac x^4e^-x1+e^-x dx$
Converting to a geometric series: $frac y1+y = sum_limitsn=1^infty (-y)^n$
$int x^4sum_limitsn=1^infty (-1)^ne^-nx dx$
considering just one term $int x^4e^-nx = (frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
$sum_limitsn=1^infty(-1)^n(frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
It is worth noting
$frac1Gamma(s) int_0^infty frac x^s-1e^x -1 dx = zeta(s)$
answered Jul 20 at 20:41
Doug M
39.2k31749
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$intfrac x^41+e^x dx\
intfrac x^4e^-x1+e^-x dx$
Converting to a geometric series: $frac y1+y = sum_limitsn=1^infty (-y)^n$
$int x^4sum_limitsn=1^infty (-1)^ne^-nx dx$
considering just one term $int x^4e^-nx = (frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
$sum_limitsn=1^infty(-1)^n(frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
It is worth noting
$frac1Gamma(s) int_0^infty frac x^s-1e^x -1 dx = zeta(s)$
answered Jul 20 at 20:41
Doug M
39.2k31749
$intfrac x^41+e^x dx\
intfrac x^4e^-x1+e^-x dx$
Converting to a geometric series: $frac y1+y = sum_limitsn=1^infty (-y)^n$
$int x^4sum_limitsn=1^infty (-1)^ne^-nx dx$
considering just one term $int x^4e^-nx = (frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
$sum_limitsn=1^infty(-1)^n(frac x^4n + frac 4x^3n^2 + frac 12x^2n^2 + frac 24xn^3 + frac 24n^4) e^-nx$
It is worth noting
$frac1Gamma(s) int_0^infty frac x^s-1e^x -1 dx = zeta(s)$
answered Jul 20 at 20:41
Doug M
39.2k31749
edited Jul 20 at 20:50
answered Jul 20 at 20:41
Doug M
39.2k31749
answered Jul 20 at 20:41
Doug M
39.2k31749
answered Jul 20 at 20:41
Doug M
39.2k31749
39.2k31749
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
add a comment |Â
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
I am a big fan of this approach to integrals. Can be used to solve one of the greatest integrals of all time, namely $displaystyleint_0^inftydfracx^p-1e^x - 1 dx = zeta(p) Gamma(p)$
– packetpacket
Jul 20 at 20:50
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
looks like your ninja edit beat me to it!
– packetpacket
Jul 20 at 20:51
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
@packetpacket thx... It usually takes me 3 revisions before I am satisfied with most of my answers.
– Doug M
Jul 20 at 20:52
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
It would be nice to remember that we can exchange integral with the series because of the uniform convergence :)
– Gonzalo Benavides
Jul 20 at 21:32
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
@DougM Sir, shouldn't the geometric series be $ frac-y1+y$ instead of $fracy1+y$ ?
– William
Jul 21 at 9:23
add a comment |Â
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3
Why do you want to calculate this integral? I suspect that you actually have another problem which involves this integral.
– md2perpe
Jul 20 at 19:58
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Jul 20 at 20:00
@md2perpe Not really I'm just playing around with integrals. (Even though I'm a rookie)
– William
Jul 20 at 20:01
The integral is very similar to the one here.
– md2perpe
Jul 20 at 20:01
@md2perpe Yea I know, I did check that a while back, but it wasn't much help :(
– William
Jul 20 at 20:03