Mixing
$int_{-infty}^{infty} (x+1)(x-1)e^{-(x+1)^2(x-1)^2}$
4
$begingroup$
I am tring to solve the following integral:
$$I=int_{-infty}^infty(x+1)(x-1)e^{-(x+1)^2(x-1)^2}dx$$
So far I cannot think of a method for doing this other than numerically, And I have no proof that it has an elementary solution. Any thoughts?
integration numerical-methods
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
$endgroup$
|
show 3 more comments
4
$begingroup$
I am tring to solve the following integral:
$$I=int_{-infty}^infty(x+1)(x-1)e^{-(x+1)^2(x-1)^2}dx$$
So far I cannot think of a method for doing this other than numerically, And I have no proof that it has an elementary solution. Any thoughts?
integration numerical-methods
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
$endgroup$
1
$begingroup$
You can express the result with the Bessel function
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:11
1
$begingroup$
see here $$frac{pi left(I_{-frac{5}{4}}left(frac{1}{2}right)+ I_{frac{5}{4}}left(frac{1}{2}right)right)} {4 sqrt{e}}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:13
1
$begingroup$
Through substitution the integral should reduce to $I=int_0^infty frac{ue^{-u^2}}{sqrt{u+1}}du$, but I'm not sure if that makes it any easier.
$endgroup$
– aleden
Jan 12 at 16:14
$begingroup$
@aleden How'd you get the lower bound (0)?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 18:34
1
$begingroup$
@Cardioid_Ass_22 yes you are correct, my mistake. It should be $I=int_{-1}^infty frac{ue^{-u^2}}{sqrt{u+1}}du$
$endgroup$
– aleden
Jan 12 at 20:19
|
show 3 more comments
4
4
4
$begingroup$
I am tring to solve the following integral:
$$I=int_{-infty}^infty(x+1)(x-1)e^{-(x+1)^2(x-1)^2}dx$$
So far I cannot think of a method for doing this other than numerically, And I have no proof that it has an elementary solution. Any thoughts?
integration numerical-methods
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
$endgroup$
I am tring to solve the following integral:
$$I=int_{-infty}^infty(x+1)(x-1)e^{-(x+1)^2(x-1)^2}dx$$
So far I cannot think of a method for doing this other than numerically, And I have no proof that it has an elementary solution. Any thoughts?
integration numerical-methods
integration numerical-methods
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
asked Jan 12 at 14:07
Henry LeeHenry Lee
1,966219
1,966219
1
$begingroup$
You can express the result with the Bessel function
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:11
1
$begingroup$
see here $$frac{pi left(I_{-frac{5}{4}}left(frac{1}{2}right)+ I_{frac{5}{4}}left(frac{1}{2}right)right)} {4 sqrt{e}}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:13
1
$begingroup$
Through substitution the integral should reduce to $I=int_0^infty frac{ue^{-u^2}}{sqrt{u+1}}du$, but I'm not sure if that makes it any easier.
$endgroup$
– aleden
Jan 12 at 16:14
$begingroup$
@aleden How'd you get the lower bound (0)?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 18:34
1
$begingroup$
@Cardioid_Ass_22 yes you are correct, my mistake. It should be $I=int_{-1}^infty frac{ue^{-u^2}}{sqrt{u+1}}du$
$endgroup$
– aleden
Jan 12 at 20:19
|
show 3 more comments
1
$begingroup$
You can express the result with the Bessel function
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:11
1
$begingroup$
see here $$frac{pi left(I_{-frac{5}{4}}left(frac{1}{2}right)+ I_{frac{5}{4}}left(frac{1}{2}right)right)} {4 sqrt{e}}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:13
1
$begingroup$
Through substitution the integral should reduce to $I=int_0^infty frac{ue^{-u^2}}{sqrt{u+1}}du$, but I'm not sure if that makes it any easier.
$endgroup$
– aleden
Jan 12 at 16:14
$begingroup$
@aleden How'd you get the lower bound (0)?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 18:34
1
$begingroup$
@Cardioid_Ass_22 yes you are correct, my mistake. It should be $I=int_{-1}^infty frac{ue^{-u^2}}{sqrt{u+1}}du$
$endgroup$
– aleden
Jan 12 at 20:19
1
1
$begingroup$
You can express the result with the Bessel function
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:11
$begingroup$
You can express the result with the Bessel function
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:11
1
1
$begingroup$
see here $$frac{pi left(I_{-frac{5}{4}}left(frac{1}{2}right)+ I_{frac{5}{4}}left(frac{1}{2}right)right)} {4 sqrt{e}}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:13
$begingroup$
see here $$frac{pi left(I_{-frac{5}{4}}left(frac{1}{2}right)+ I_{frac{5}{4}}left(frac{1}{2}right)right)} {4 sqrt{e}}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:13
1
1
$begingroup$
Through substitution the integral should reduce to $I=int_0^infty frac{ue^{-u^2}}{sqrt{u+1}}du$, but I'm not sure if that makes it any easier.
$endgroup$
– aleden
Jan 12 at 16:14
$begingroup$
Through substitution the integral should reduce to $I=int_0^infty frac{ue^{-u^2}}{sqrt{u+1}}du$, but I'm not sure if that makes it any easier.
$endgroup$
– aleden
Jan 12 at 16:14
$begingroup$
@aleden How'd you get the lower bound (0)?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 18:34
$begingroup$
@aleden How'd you get the lower bound (0)?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 18:34
1
1
$begingroup$
@Cardioid_Ass_22 yes you are correct, my mistake. It should be $I=int_{-1}^infty frac{ue^{-u^2}}{sqrt{u+1}}du$
$endgroup$
– aleden
Jan 12 at 20:19
$begingroup$
@Cardioid_Ass_22 yes you are correct, my mistake. It should be $I=int_{-1}^infty frac{ue^{-u^2}}{sqrt{u+1}}du$
$endgroup$
– aleden
Jan 12 at 20:19
|
show 3 more comments
0
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1
$begingroup$
You can express the result with the Bessel function
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:11
1
$begingroup$
see here $$frac{pi left(I_{-frac{5}{4}}left(frac{1}{2}right)+ I_{frac{5}{4}}left(frac{1}{2}right)right)} {4 sqrt{e}}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 14:13
1
$begingroup$
Through substitution the integral should reduce to $I=int_0^infty frac{ue^{-u^2}}{sqrt{u+1}}du$, but I'm not sure if that makes it any easier.
$endgroup$
– aleden
Jan 12 at 16:14
$begingroup$
@aleden How'd you get the lower bound (0)?
$endgroup$
– Cardioid_Ass_22
Jan 12 at 18:34
1
$begingroup$
@Cardioid_Ass_22 yes you are correct, my mistake. It should be $I=int_{-1}^infty frac{ue^{-u^2}}{sqrt{u+1}}du$
$endgroup$
– aleden
Jan 12 at 20:19