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




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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















up vote
5
down vote

favorite
1












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




share|cite|improve this question





















  • 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













up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





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




share|cite|improve this question













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





share|cite|improve this question












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edited Aug 6 at 3:34
























asked Aug 6 at 2:22









D. Shelly

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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

















  • 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











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?






share|cite|improve this answer





















  • 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










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1 Answer
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active

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active

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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?






share|cite|improve this answer





















  • 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














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?






share|cite|improve this answer





















  • 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












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?






share|cite|improve this answer













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?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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
















  • 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












 

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