Mixing
Indefinite Integral for $cos x/(1+x^2)$
$begingroup$
I have been working on the indefinite integral of $cos x/(1+x^2)$.
$$ intfrac{cos x}{1+x^2};dxtext{ or } intfrac{sin x}{1+x^2};dx $$
are they unsolvable(impossible to solve) or is there a way to solve them even by approximation?
Thank you very much.
analysis indefinite-integrals
edited Mar 27 '13 at 10:31
user642796
44.9k564119
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
$endgroup$
|
show 3 more comments
$begingroup$
I have been working on the indefinite integral of $cos x/(1+x^2)$.
$$ intfrac{cos x}{1+x^2};dxtext{ or } intfrac{sin x}{1+x^2};dx $$
are they unsolvable(impossible to solve) or is there a way to solve them even by approximation?
Thank you very much.
analysis indefinite-integrals
edited Mar 27 '13 at 10:31
user642796
44.9k564119
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
$endgroup$
2
$begingroup$
wolfram alpha provides a solution in terms of sine and cosine-integrals.
$endgroup$
– Fabian
Apr 9 '12 at 16:20
$begingroup$
Why is the word "undefined" in the title?
$endgroup$
– Michael Hardy
Apr 9 '12 at 16:29
1
$begingroup$
I guess there is no explicit formula for the indefinite integral. I know and estimate $$ int_{0}^{pi/2}frac{cos x}{1+x^2};dxge int_{0}^{pi/2}frac{sin x}{1+x^2};dx $$
$endgroup$
– Sunni
Apr 9 '12 at 16:59
1
$begingroup$
I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?
$endgroup$
– Fabian
Apr 9 '12 at 17:27
2
$begingroup$
Defacing your questions is quite frowned upon; please don't do this.
$endgroup$
– user642796
Mar 27 '13 at 10:33
|
show 3 more comments
$begingroup$
I have been working on the indefinite integral of $cos x/(1+x^2)$.
$$ intfrac{cos x}{1+x^2};dxtext{ or } intfrac{sin x}{1+x^2};dx $$
are they unsolvable(impossible to solve) or is there a way to solve them even by approximation?
Thank you very much.
analysis indefinite-integrals
edited Mar 27 '13 at 10:31
user642796
44.9k564119
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
$endgroup$
I have been working on the indefinite integral of $cos x/(1+x^2)$.
$$ intfrac{cos x}{1+x^2};dxtext{ or } intfrac{sin x}{1+x^2};dx $$
are they unsolvable(impossible to solve) or is there a way to solve them even by approximation?
Thank you very much.
analysis indefinite-integrals
analysis indefinite-integrals
edited Mar 27 '13 at 10:31
user642796
44.9k564119
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
edited Mar 27 '13 at 10:31
user642796
44.9k564119
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
edited Mar 27 '13 at 10:31
user642796
44.9k564119
edited Mar 27 '13 at 10:31
user642796
44.9k564119
edited Mar 27 '13 at 10:31
user642796
44.9k564119
44.9k564119
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
asked Apr 9 '12 at 16:18
Bilgis77Bilgis77
146226
146226
2
$begingroup$
wolfram alpha provides a solution in terms of sine and cosine-integrals.
$endgroup$
– Fabian
Apr 9 '12 at 16:20
$begingroup$
Why is the word "undefined" in the title?
$endgroup$
– Michael Hardy
Apr 9 '12 at 16:29
1
$begingroup$
I guess there is no explicit formula for the indefinite integral. I know and estimate $$ int_{0}^{pi/2}frac{cos x}{1+x^2};dxge int_{0}^{pi/2}frac{sin x}{1+x^2};dx $$
$endgroup$
– Sunni
Apr 9 '12 at 16:59
1
$begingroup$
I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?
$endgroup$
– Fabian
Apr 9 '12 at 17:27
2
$begingroup$
Defacing your questions is quite frowned upon; please don't do this.
$endgroup$
– user642796
Mar 27 '13 at 10:33
|
show 3 more comments
2
$begingroup$
wolfram alpha provides a solution in terms of sine and cosine-integrals.
$endgroup$
– Fabian
Apr 9 '12 at 16:20
$begingroup$
Why is the word "undefined" in the title?
$endgroup$
– Michael Hardy
Apr 9 '12 at 16:29
1
$begingroup$
I guess there is no explicit formula for the indefinite integral. I know and estimate $$ int_{0}^{pi/2}frac{cos x}{1+x^2};dxge int_{0}^{pi/2}frac{sin x}{1+x^2};dx $$
$endgroup$
– Sunni
Apr 9 '12 at 16:59
1
$begingroup$
I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?
$endgroup$
– Fabian
Apr 9 '12 at 17:27
2
$begingroup$
Defacing your questions is quite frowned upon; please don't do this.
$endgroup$
– user642796
Mar 27 '13 at 10:33
2
2
$begingroup$
wolfram alpha provides a solution in terms of sine and cosine-integrals.
$endgroup$
– Fabian
Apr 9 '12 at 16:20
$begingroup$
wolfram alpha provides a solution in terms of sine and cosine-integrals.
$endgroup$
– Fabian
Apr 9 '12 at 16:20
$begingroup$
Why is the word "undefined" in the title?
$endgroup$
– Michael Hardy
Apr 9 '12 at 16:29
$begingroup$
Why is the word "undefined" in the title?
$endgroup$
– Michael Hardy
Apr 9 '12 at 16:29
1
1
$begingroup$
I guess there is no explicit formula for the indefinite integral. I know and estimate $$ int_{0}^{pi/2}frac{cos x}{1+x^2};dxge int_{0}^{pi/2}frac{sin x}{1+x^2};dx $$
$endgroup$
– Sunni
Apr 9 '12 at 16:59
$begingroup$
I guess there is no explicit formula for the indefinite integral. I know and estimate $$ int_{0}^{pi/2}frac{cos x}{1+x^2};dxge int_{0}^{pi/2}frac{sin x}{1+x^2};dx $$
$endgroup$
– Sunni
Apr 9 '12 at 16:59
1
1
$begingroup$
I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?
$endgroup$
– Fabian
Apr 9 '12 at 17:27
$begingroup$
I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?
$endgroup$
– Fabian
Apr 9 '12 at 17:27
2
2
$begingroup$
Defacing your questions is quite frowned upon; please don't do this.
$endgroup$
– user642796
Mar 27 '13 at 10:33
$begingroup$
Defacing your questions is quite frowned upon; please don't do this.
$endgroup$
– user642796
Mar 27 '13 at 10:33
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
There is no elementary antiderivative for either of those.
It's actually easier to deal with $e^{ix}/(1+x^2)$. As a corollary of a theorem of Liouville, if $f e^g$ has an elementary antiderivative, where $f$ and $g$ are rational functions and $g$ is not constant, then it has an antiderivative of the form $h e^g$ where $h$ is a rational function. For this to be an antiderivative of $f e^g$, what we need is $h' + h g' = f$.
Now with $f = 1/(1+x^2)$ and $g = ix$, the condition is $h' + i h = 1/(1+x^2)$. The right side has a pole of order $1$ at $x=i$. In order for the left side to have a pole there,
$h$ must have a pole there, but wherever $h$ has a pole of order $k$, $h'$ has a pole of order $k+1$, so the left side can never have a pole of order $1$.
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
$endgroup$
add a comment |
$begingroup$
$intdfrac{sin x}{1+x^2}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftydfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(sumlimits_{n=0}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{sinh1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=dfrac{sinh1ln(x^2+1)}{2}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
There is no elementary antiderivative for either of those.
It's actually easier to deal with $e^{ix}/(1+x^2)$. As a corollary of a theorem of Liouville, if $f e^g$ has an elementary antiderivative, where $f$ and $g$ are rational functions and $g$ is not constant, then it has an antiderivative of the form $h e^g$ where $h$ is a rational function. For this to be an antiderivative of $f e^g$, what we need is $h' + h g' = f$.
Now with $f = 1/(1+x^2)$ and $g = ix$, the condition is $h' + i h = 1/(1+x^2)$. The right side has a pole of order $1$ at $x=i$. In order for the left side to have a pole there,
$h$ must have a pole there, but wherever $h$ has a pole of order $k$, $h'$ has a pole of order $k+1$, so the left side can never have a pole of order $1$.
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
$endgroup$
add a comment |
$begingroup$
There is no elementary antiderivative for either of those.
It's actually easier to deal with $e^{ix}/(1+x^2)$. As a corollary of a theorem of Liouville, if $f e^g$ has an elementary antiderivative, where $f$ and $g$ are rational functions and $g$ is not constant, then it has an antiderivative of the form $h e^g$ where $h$ is a rational function. For this to be an antiderivative of $f e^g$, what we need is $h' + h g' = f$.
Now with $f = 1/(1+x^2)$ and $g = ix$, the condition is $h' + i h = 1/(1+x^2)$. The right side has a pole of order $1$ at $x=i$. In order for the left side to have a pole there,
$h$ must have a pole there, but wherever $h$ has a pole of order $k$, $h'$ has a pole of order $k+1$, so the left side can never have a pole of order $1$.
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
$endgroup$
add a comment |
$begingroup$
There is no elementary antiderivative for either of those.
It's actually easier to deal with $e^{ix}/(1+x^2)$. As a corollary of a theorem of Liouville, if $f e^g$ has an elementary antiderivative, where $f$ and $g$ are rational functions and $g$ is not constant, then it has an antiderivative of the form $h e^g$ where $h$ is a rational function. For this to be an antiderivative of $f e^g$, what we need is $h' + h g' = f$.
Now with $f = 1/(1+x^2)$ and $g = ix$, the condition is $h' + i h = 1/(1+x^2)$. The right side has a pole of order $1$ at $x=i$. In order for the left side to have a pole there,
$h$ must have a pole there, but wherever $h$ has a pole of order $k$, $h'$ has a pole of order $k+1$, so the left side can never have a pole of order $1$.
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
$endgroup$
There is no elementary antiderivative for either of those.
It's actually easier to deal with $e^{ix}/(1+x^2)$. As a corollary of a theorem of Liouville, if $f e^g$ has an elementary antiderivative, where $f$ and $g$ are rational functions and $g$ is not constant, then it has an antiderivative of the form $h e^g$ where $h$ is a rational function. For this to be an antiderivative of $f e^g$, what we need is $h' + h g' = f$.
Now with $f = 1/(1+x^2)$ and $g = ix$, the condition is $h' + i h = 1/(1+x^2)$. The right side has a pole of order $1$ at $x=i$. In order for the left side to have a pole there,
$h$ must have a pole there, but wherever $h$ has a pole of order $k$, $h'$ has a pole of order $k+1$, so the left side can never have a pole of order $1$.
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
answered May 9 '12 at 21:39
Robert IsraelRobert Israel
328k23216469
328k23216469
add a comment |
add a comment |
$begingroup$
$intdfrac{sin x}{1+x^2}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftydfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(sumlimits_{n=0}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{sinh1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=dfrac{sinh1ln(x^2+1)}{2}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
$endgroup$
add a comment |
$begingroup$
$intdfrac{sin x}{1+x^2}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftydfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(sumlimits_{n=0}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{sinh1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=dfrac{sinh1ln(x^2+1)}{2}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
$endgroup$
add a comment |
$begingroup$
$intdfrac{sin x}{1+x^2}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftydfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(sumlimits_{n=0}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{sinh1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=dfrac{sinh1ln(x^2+1)}{2}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
$endgroup$
$intdfrac{sin x}{1+x^2}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$
$=intsumlimits_{n=0}^inftydfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftydfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=intsumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{1}{2(x^2+1)}+sumlimits_{n=1}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(sumlimits_{n=0}^inftydfrac{1}{2(2n+1)!(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=intleft(dfrac{sinh1}{2(x^2+1)}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}right)d(x^2+1)$
$=dfrac{sinh1ln(x^2+1)}{2}+sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
answered May 29 '15 at 15:47
Harry PeterHarry Peter
5,48911439
5,48911439
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2
$begingroup$
wolfram alpha provides a solution in terms of sine and cosine-integrals.
$endgroup$
– Fabian
Apr 9 '12 at 16:20
$begingroup$
Why is the word "undefined" in the title?
$endgroup$
– Michael Hardy
Apr 9 '12 at 16:29
1
$begingroup$
I guess there is no explicit formula for the indefinite integral. I know and estimate $$ int_{0}^{pi/2}frac{cos x}{1+x^2};dxge int_{0}^{pi/2}frac{sin x}{1+x^2};dx $$
$endgroup$
– Sunni
Apr 9 '12 at 16:59
1
$begingroup$
I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?
$endgroup$
– Fabian
Apr 9 '12 at 17:27
2
$begingroup$
Defacing your questions is quite frowned upon; please don't do this.
$endgroup$
– user642796
Mar 27 '13 at 10:33