Help with a very hard integral [closed]
$begingroup$
$$
int^{infty}_{1} Bigl(,Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+xBigr|-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x=
int^{infty}_{1} Bigl(3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+x-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x$$
I literally have no idea on how to do this, perhaps through complex analysis or polar coordinates(?). Also I only want an exact answer.
Thanks!
Edit:
Thanks to metamorphy for pointing out that I don't need the first absolute value.
integration definite-integrals improper-integrals
edited Dec 20 '18 at 13:13
egreg
182k1485204
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
$endgroup$
closed as off-topic by amWhy, Leucippus, Arjang, Saad, Cesareo Dec 21 '18 at 9:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Arjang, Saad, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$
int^{infty}_{1} Bigl(,Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+xBigr|-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x=
int^{infty}_{1} Bigl(3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+x-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x$$
I literally have no idea on how to do this, perhaps through complex analysis or polar coordinates(?). Also I only want an exact answer.
Thanks!
Edit:
Thanks to metamorphy for pointing out that I don't need the first absolute value.
integration definite-integrals improper-integrals
edited Dec 20 '18 at 13:13
egreg
182k1485204
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
$endgroup$
closed as off-topic by amWhy, Leucippus, Arjang, Saad, Cesareo Dec 21 '18 at 9:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Arjang, Saad, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You want the nature or the value ?
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 12:55
$begingroup$
You can remove the second absolute value as well, as the inside is always negative, multiply by a negative sign after removing and simplify the expression.
$endgroup$
– Uday Khanna
Dec 20 '18 at 13:00
$begingroup$
@haman_Abdallah What do you mean by "nature"?
$endgroup$
– Lars Aadnesen
Dec 20 '18 at 13:01
$begingroup$
Welcome to the site ! I am sorry : this is not a very hard integral ! It is just a pure nightmare. Cheers.
$endgroup$
– Claude Leibovici
Dec 20 '18 at 14:06
add a comment |
$begingroup$
$$
int^{infty}_{1} Bigl(,Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+xBigr|-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x=
int^{infty}_{1} Bigl(3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+x-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x$$
I literally have no idea on how to do this, perhaps through complex analysis or polar coordinates(?). Also I only want an exact answer.
Thanks!
Edit:
Thanks to metamorphy for pointing out that I don't need the first absolute value.
integration definite-integrals improper-integrals
edited Dec 20 '18 at 13:13
egreg
182k1485204
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
$endgroup$
$$
int^{infty}_{1} Bigl(,Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+xBigr|-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x=
int^{infty}_{1} Bigl(3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}+x-Bigl|3mathrm{e}^{-tfrac{x^2}{sqrt{x-1}}}-xBigr|,Bigr),mathrm{d}x$$
I literally have no idea on how to do this, perhaps through complex analysis or polar coordinates(?). Also I only want an exact answer.
Thanks!
Edit:
Thanks to metamorphy for pointing out that I don't need the first absolute value.
integration definite-integrals improper-integrals
integration definite-integrals improper-integrals
edited Dec 20 '18 at 13:13
egreg
182k1485204
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
edited Dec 20 '18 at 13:13
egreg
182k1485204
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
edited Dec 20 '18 at 13:13
egreg
182k1485204
edited Dec 20 '18 at 13:13
egreg
182k1485204
edited Dec 20 '18 at 13:13
egreg
182k1485204
182k1485204
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
asked Dec 20 '18 at 12:45
Lars AadnesenLars Aadnesen
11
11
closed as off-topic by amWhy, Leucippus, Arjang, Saad, Cesareo Dec 21 '18 at 9:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Arjang, Saad, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Leucippus, Arjang, Saad, Cesareo Dec 21 '18 at 9:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Arjang, Saad, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You want the nature or the value ?
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 12:55
$begingroup$
You can remove the second absolute value as well, as the inside is always negative, multiply by a negative sign after removing and simplify the expression.
$endgroup$
– Uday Khanna
Dec 20 '18 at 13:00
$begingroup$
@haman_Abdallah What do you mean by "nature"?
$endgroup$
– Lars Aadnesen
Dec 20 '18 at 13:01
$begingroup$
Welcome to the site ! I am sorry : this is not a very hard integral ! It is just a pure nightmare. Cheers.
$endgroup$
– Claude Leibovici
Dec 20 '18 at 14:06
add a comment |
$begingroup$
You want the nature or the value ?
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 12:55
$begingroup$
You can remove the second absolute value as well, as the inside is always negative, multiply by a negative sign after removing and simplify the expression.
$endgroup$
– Uday Khanna
Dec 20 '18 at 13:00
$begingroup$
@haman_Abdallah What do you mean by "nature"?
$endgroup$
– Lars Aadnesen
Dec 20 '18 at 13:01
$begingroup$
Welcome to the site ! I am sorry : this is not a very hard integral ! It is just a pure nightmare. Cheers.
$endgroup$
– Claude Leibovici
Dec 20 '18 at 14:06
$begingroup$
You want the nature or the value ?
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 12:55
$begingroup$
You want the nature or the value ?
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 12:55
$begingroup$
You can remove the second absolute value as well, as the inside is always negative, multiply by a negative sign after removing and simplify the expression.
$endgroup$
– Uday Khanna
Dec 20 '18 at 13:00
$begingroup$
You can remove the second absolute value as well, as the inside is always negative, multiply by a negative sign after removing and simplify the expression.
$endgroup$
– Uday Khanna
Dec 20 '18 at 13:00
$begingroup$
@haman_Abdallah What do you mean by "nature"?
$endgroup$
– Lars Aadnesen
Dec 20 '18 at 13:01
$begingroup$
@haman_Abdallah What do you mean by "nature"?
$endgroup$
– Lars Aadnesen
Dec 20 '18 at 13:01
$begingroup$
Welcome to the site ! I am sorry : this is not a very hard integral ! It is just a pure nightmare. Cheers.
$endgroup$
– Claude Leibovici
Dec 20 '18 at 14:06
$begingroup$
Welcome to the site ! I am sorry : this is not a very hard integral ! It is just a pure nightmare. Cheers.
$endgroup$
– Claude Leibovici
Dec 20 '18 at 14:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The first step is proving $3exp -frac{x^2}{sqrt{x-1}}<x$ for all $xge 1$. (Judging by this graph, you just need to check its derivative is negative.) So your integral is $$int_1^infty left(3exp -frac{x^2}{sqrt{x-1}}+x+3exp -frac{x^2}{sqrt{x-1}}-xright)dx=6int_1^inftyexp -frac{x^2}{sqrt{x-1}} dx.$$However, there doesn't seem to be a non-numerical method for this.
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first step is proving $3exp -frac{x^2}{sqrt{x-1}}<x$ for all $xge 1$. (Judging by this graph, you just need to check its derivative is negative.) So your integral is $$int_1^infty left(3exp -frac{x^2}{sqrt{x-1}}+x+3exp -frac{x^2}{sqrt{x-1}}-xright)dx=6int_1^inftyexp -frac{x^2}{sqrt{x-1}} dx.$$However, there doesn't seem to be a non-numerical method for this.
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
The first step is proving $3exp -frac{x^2}{sqrt{x-1}}<x$ for all $xge 1$. (Judging by this graph, you just need to check its derivative is negative.) So your integral is $$int_1^infty left(3exp -frac{x^2}{sqrt{x-1}}+x+3exp -frac{x^2}{sqrt{x-1}}-xright)dx=6int_1^inftyexp -frac{x^2}{sqrt{x-1}} dx.$$However, there doesn't seem to be a non-numerical method for this.
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
The first step is proving $3exp -frac{x^2}{sqrt{x-1}}<x$ for all $xge 1$. (Judging by this graph, you just need to check its derivative is negative.) So your integral is $$int_1^infty left(3exp -frac{x^2}{sqrt{x-1}}+x+3exp -frac{x^2}{sqrt{x-1}}-xright)dx=6int_1^inftyexp -frac{x^2}{sqrt{x-1}} dx.$$However, there doesn't seem to be a non-numerical method for this.
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
$endgroup$
The first step is proving $3exp -frac{x^2}{sqrt{x-1}}<x$ for all $xge 1$. (Judging by this graph, you just need to check its derivative is negative.) So your integral is $$int_1^infty left(3exp -frac{x^2}{sqrt{x-1}}+x+3exp -frac{x^2}{sqrt{x-1}}-xright)dx=6int_1^inftyexp -frac{x^2}{sqrt{x-1}} dx.$$However, there doesn't seem to be a non-numerical method for this.
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
answered Dec 20 '18 at 13:01
J.G.J.G.
27.1k22843
27.1k22843
add a comment |
add a comment |
$begingroup$
You want the nature or the value ?
$endgroup$
– hamam_Abdallah
Dec 20 '18 at 12:55
$begingroup$
You can remove the second absolute value as well, as the inside is always negative, multiply by a negative sign after removing and simplify the expression.
$endgroup$
– Uday Khanna
Dec 20 '18 at 13:00
$begingroup$
@haman_Abdallah What do you mean by "nature"?
$endgroup$
– Lars Aadnesen
Dec 20 '18 at 13:01
$begingroup$
Welcome to the site ! I am sorry : this is not a very hard integral ! It is just a pure nightmare. Cheers.
$endgroup$
– Claude Leibovici
Dec 20 '18 at 14:06