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Generalized Infinite Integration by Parts
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During my studies in calc 2, I became fascinated by the integral $int e^-x^2dx$ after hearing from the professor that it has no elementary function as its integral. I came up with an interesting technique to try the integral by using Integration by Parts like so: $$int e^-x^2dx=xe^-x^2+2int x^2e^-x^2dx $$ $$int e^-x^2dx=xe^-x^2+frac 2 3x^3e^-x^2+int x^4e^-x^2dx$$
and continuing this until the pattern became obvious and I came up with the following equation: $$int e^-x^2dx=e^-x^2sum_i=0^infty
frac 2^i (2i+1)!!x^2i+1+C$$ where $(2i+1)!!$ denotes the double factorial $(2n+1)(2n-1)(2n-3)...(2 text or 1)$. From here, I thought about a generalized case for any infinitely differentiable function $f(x)$ and for an argument raised to any power $n$. $$int f(x^n)dx $$ $$u=f(x^n), du=nx^n-1f'(x^n), dv=dx, v=x$$ $$int f(x^n)dx=xf(x^n)-nint x^nf'(x^n)dx$$ and eventually: $$=sum_i=0^inftyfrac (-n)^ix^ni+1 (ni+1)(n(i-1)+1)...(n+1)(1)f^(i)(x^n)+C$$ My question is twofold. First, did I make any glaring mistakes, and second, is this particular formula useful or novel?
calculus integration sequences-and-series
asked Aug 6 at 2:22
D. Shelly
263
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up vote
5
down vote
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During my studies in calc 2, I became fascinated by the integral $int e^-x^2dx$ after hearing from the professor that it has no elementary function as its integral. I came up with an interesting technique to try the integral by using Integration by Parts like so: $$int e^-x^2dx=xe^-x^2+2int x^2e^-x^2dx $$ $$int e^-x^2dx=xe^-x^2+frac 2 3x^3e^-x^2+int x^4e^-x^2dx$$
and continuing this until the pattern became obvious and I came up with the following equation: $$int e^-x^2dx=e^-x^2sum_i=0^infty
frac 2^i (2i+1)!!x^2i+1+C$$ where $(2i+1)!!$ denotes the double factorial $(2n+1)(2n-1)(2n-3)...(2 text or 1)$. From here, I thought about a generalized case for any infinitely differentiable function $f(x)$ and for an argument raised to any power $n$. $$int f(x^n)dx $$ $$u=f(x^n), du=nx^n-1f'(x^n), dv=dx, v=x$$ $$int f(x^n)dx=xf(x^n)-nint x^nf'(x^n)dx$$ and eventually: $$=sum_i=0^inftyfrac (-n)^ix^ni+1 (ni+1)(n(i-1)+1)...(n+1)(1)f^(i)(x^n)+C$$ My question is twofold. First, did I make any glaring mistakes, and second, is this particular formula useful or novel?
calculus integration sequences-and-series
asked Aug 6 at 2:22
D. Shelly
263
repeated integration by parts is a way we get asymptotic expansions store.doverpublications.com/0486650820.html
– Will Jagy
Aug 6 at 2:27
Your $du$ is off. Should be $nx^n-1f'(x^n)$.
– Randall
Aug 6 at 2:28
I don't understand why you're doing what you're doing when the taylor series of $e^x$ is known and you just integrate that.
– RHowe
Aug 6 at 2:28
3
@Geronimo I don't understand why anyone interested in math would not do this, if it occurred to them.
– Cory Griffith
Aug 6 at 2:30
1
I would also like to encourage OP—who, bear in mind, is a Calc 2 student—to continue to investigate and ponder questions about things seen in class.
– Randall
Aug 6 at 2:46
 |Â
show 5 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
During my studies in calc 2, I became fascinated by the integral $int e^-x^2dx$ after hearing from the professor that it has no elementary function as its integral. I came up with an interesting technique to try the integral by using Integration by Parts like so: $$int e^-x^2dx=xe^-x^2+2int x^2e^-x^2dx $$ $$int e^-x^2dx=xe^-x^2+frac 2 3x^3e^-x^2+int x^4e^-x^2dx$$
and continuing this until the pattern became obvious and I came up with the following equation: $$int e^-x^2dx=e^-x^2sum_i=0^infty
frac 2^i (2i+1)!!x^2i+1+C$$ where $(2i+1)!!$ denotes the double factorial $(2n+1)(2n-1)(2n-3)...(2 text or 1)$. From here, I thought about a generalized case for any infinitely differentiable function $f(x)$ and for an argument raised to any power $n$. $$int f(x^n)dx $$ $$u=f(x^n), du=nx^n-1f'(x^n), dv=dx, v=x$$ $$int f(x^n)dx=xf(x^n)-nint x^nf'(x^n)dx$$ and eventually: $$=sum_i=0^inftyfrac (-n)^ix^ni+1 (ni+1)(n(i-1)+1)...(n+1)(1)f^(i)(x^n)+C$$ My question is twofold. First, did I make any glaring mistakes, and second, is this particular formula useful or novel?
calculus integration sequences-and-series
asked Aug 6 at 2:22
D. Shelly
263
During my studies in calc 2, I became fascinated by the integral $int e^-x^2dx$ after hearing from the professor that it has no elementary function as its integral. I came up with an interesting technique to try the integral by using Integration by Parts like so: $$int e^-x^2dx=xe^-x^2+2int x^2e^-x^2dx $$ $$int e^-x^2dx=xe^-x^2+frac 2 3x^3e^-x^2+int x^4e^-x^2dx$$
and continuing this until the pattern became obvious and I came up with the following equation: $$int e^-x^2dx=e^-x^2sum_i=0^infty
frac 2^i (2i+1)!!x^2i+1+C$$ where $(2i+1)!!$ denotes the double factorial $(2n+1)(2n-1)(2n-3)...(2 text or 1)$. From here, I thought about a generalized case for any infinitely differentiable function $f(x)$ and for an argument raised to any power $n$. $$int f(x^n)dx $$ $$u=f(x^n), du=nx^n-1f'(x^n), dv=dx, v=x$$ $$int f(x^n)dx=xf(x^n)-nint x^nf'(x^n)dx$$ and eventually: $$=sum_i=0^inftyfrac (-n)^ix^ni+1 (ni+1)(n(i-1)+1)...(n+1)(1)f^(i)(x^n)+C$$ My question is twofold. First, did I make any glaring mistakes, and second, is this particular formula useful or novel?
calculus integration sequences-and-series
asked Aug 6 at 2:22
D. Shelly
263
edited Aug 6 at 3:34
asked Aug 6 at 2:22
D. Shelly
263
asked Aug 6 at 2:22
D. Shelly
263
asked Aug 6 at 2:22
D. Shelly
263
263
repeated integration by parts is a way we get asymptotic expansions store.doverpublications.com/0486650820.html
– Will Jagy
Aug 6 at 2:27
Your $du$ is off. Should be $nx^n-1f'(x^n)$.
– Randall
Aug 6 at 2:28
I don't understand why you're doing what you're doing when the taylor series of $e^x$ is known and you just integrate that.
– RHowe
Aug 6 at 2:28
3
@Geronimo I don't understand why anyone interested in math would not do this, if it occurred to them.
– Cory Griffith
Aug 6 at 2:30
1
I would also like to encourage OP—who, bear in mind, is a Calc 2 student—to continue to investigate and ponder questions about things seen in class.
– Randall
Aug 6 at 2:46
 |Â
show 5 more comments
repeated integration by parts is a way we get asymptotic expansions store.doverpublications.com/0486650820.html
– Will Jagy
Aug 6 at 2:27
Your $du$ is off. Should be $nx^n-1f'(x^n)$.
– Randall
Aug 6 at 2:28
I don't understand why you're doing what you're doing when the taylor series of $e^x$ is known and you just integrate that.
– RHowe
Aug 6 at 2:28
3
@Geronimo I don't understand why anyone interested in math would not do this, if it occurred to them.
– Cory Griffith
Aug 6 at 2:30
1
I would also like to encourage OP—who, bear in mind, is a Calc 2 student—to continue to investigate and ponder questions about things seen in class.
– Randall
Aug 6 at 2:46
repeated integration by parts is a way we get asymptotic expansions store.doverpublications.com/0486650820.html
– Will Jagy
Aug 6 at 2:27
repeated integration by parts is a way we get asymptotic expansions store.doverpublications.com/0486650820.html
– Will Jagy
Aug 6 at 2:27
Your $du$ is off. Should be $nx^n-1f'(x^n)$.
– Randall
Aug 6 at 2:28
Your $du$ is off. Should be $nx^n-1f'(x^n)$.
– Randall
Aug 6 at 2:28
I don't understand why you're doing what you're doing when the taylor series of $e^x$ is known and you just integrate that.
– RHowe
Aug 6 at 2:28
I don't understand why you're doing what you're doing when the taylor series of $e^x$ is known and you just integrate that.
– RHowe
Aug 6 at 2:28
3
3
@Geronimo I don't understand why anyone interested in math would not do this, if it occurred to them.
– Cory Griffith
Aug 6 at 2:30
@Geronimo I don't understand why anyone interested in math would not do this, if it occurred to them.
– Cory Griffith
Aug 6 at 2:30
1
1
I would also like to encourage OP—who, bear in mind, is a Calc 2 student—to continue to investigate and ponder questions about things seen in class.
– Randall
Aug 6 at 2:46
I would also like to encourage OP—who, bear in mind, is a Calc 2 student—to continue to investigate and ponder questions about things seen in class.
– Randall
Aug 6 at 2:46
 |Â
show 5 more comments
1 Answer
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Please check your $du$ again in $$ u=f(x^n), du=nx^n-1f(x^n), dv=dx, v=x$$
Also check this formula $$ int f(x^n)dx=xf(x^n)-nint x^nf(x^n)dx$$
What is the advantage of your method in comparison with integrating the Taylor series of a function?
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Please check your $du$ again in $$ u=f(x^n), du=nx^n-1f(x^n), dv=dx, v=x$$
Also check this formula $$ int f(x^n)dx=xf(x^n)-nint x^nf(x^n)dx$$
What is the advantage of your method in comparison with integrating the Taylor series of a function?
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
add a comment |Â
up vote
0
down vote
Please check your $du$ again in $$ u=f(x^n), du=nx^n-1f(x^n), dv=dx, v=x$$
Also check this formula $$ int f(x^n)dx=xf(x^n)-nint x^nf(x^n)dx$$
What is the advantage of your method in comparison with integrating the Taylor series of a function?
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Please check your $du$ again in $$ u=f(x^n), du=nx^n-1f(x^n), dv=dx, v=x$$
Also check this formula $$ int f(x^n)dx=xf(x^n)-nint x^nf(x^n)dx$$
What is the advantage of your method in comparison with integrating the Taylor series of a function?
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
Please check your $du$ again in $$ u=f(x^n), du=nx^n-1f(x^n), dv=dx, v=x$$
Also check this formula $$ int f(x^n)dx=xf(x^n)-nint x^nf(x^n)dx$$
What is the advantage of your method in comparison with integrating the Taylor series of a function?
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
answered Aug 6 at 2:35
Mohammad Riazi-Kermani
27.8k41852
27.8k41852
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
add a comment |Â
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thank you for pointing out my mistakes- they have been corrected. As for your question, I was just recently playing around with it on a graphing calculator and at least for the original function I was integrating it appears to converge faster than the corresponding taylor series.
– D. Shelly
Aug 7 at 16:05
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
Thanks for your attention
– Mohammad Riazi-Kermani
Aug 7 at 16:07
add a comment |Â
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repeated integration by parts is a way we get asymptotic expansions store.doverpublications.com/0486650820.html
– Will Jagy
Aug 6 at 2:27
Your $du$ is off. Should be $nx^n-1f'(x^n)$.
– Randall
Aug 6 at 2:28
I don't understand why you're doing what you're doing when the taylor series of $e^x$ is known and you just integrate that.
– RHowe
Aug 6 at 2:28
3
@Geronimo I don't understand why anyone interested in math would not do this, if it occurred to them.
– Cory Griffith
Aug 6 at 2:30
1
I would also like to encourage OP—who, bear in mind, is a Calc 2 student—to continue to investigate and ponder questions about things seen in class.
– Randall
Aug 6 at 2:46