Anyone to solve $ int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx $

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I'm trying to solve the following formula, but my calculus is quite unhandsome.

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx
$$

I tried:

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx = int_0^infty Big[ (k+1) + (n-k-1) e^y Big] ^-2 (y/alpha) e^y dy
$$

And somewhat felt that it looks related to logistic function.

Can anybody give some hints?

calculus integration definite-integrals

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edited Jul 17 at 18:31

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up vote
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1

I'm trying to solve the following formula, but my calculus is quite unhandsome.

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx
$$

I tried:

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx = int_0^infty Big[ (k+1) + (n-k-1) e^y Big] ^-2 (y/alpha) e^y dy
$$

And somewhat felt that it looks related to logistic function.

Can anybody give some hints?

calculus integration definite-integrals

share|cite|improve this question

edited Jul 17 at 18:31

Nosrati

19.8k41644

asked Jul 17 at 12:31

moreblue

1738

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I'm trying to solve the following formula, but my calculus is quite unhandsome.

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx
$$

I tried:

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx = int_0^infty Big[ (k+1) + (n-k-1) e^y Big] ^-2 (y/alpha) e^y dy
$$

And somewhat felt that it looks related to logistic function.

Can anybody give some hints?

calculus integration definite-integrals

share|cite|improve this question

edited Jul 17 at 18:31

Nosrati

19.8k41644

asked Jul 17 at 12:31

moreblue

1738

I'm trying to solve the following formula, but my calculus is quite unhandsome.

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx
$$

I tried:

$$
int_0^infty Big[ (k+1) + (n-k-1) e^alpha x Big] ^-2 alpha x e^alpha x dx = int_0^infty Big[ (k+1) + (n-k-1) e^y Big] ^-2 (y/alpha) e^y dy
$$

And somewhat felt that it looks related to logistic function.

Can anybody give some hints?

calculus integration definite-integrals

share|cite|improve this question

edited Jul 17 at 18:31

Nosrati

19.8k41644

asked Jul 17 at 12:31

moreblue

1738

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 18:31

Nosrati

19.8k41644

edited Jul 17 at 18:31

Nosrati

19.8k41644

edited Jul 17 at 18:31

Nosrati

19.8k41644

19.8k41644

asked Jul 17 at 12:31

moreblue

1738

asked Jul 17 at 12:31

moreblue

1738

asked Jul 17 at 12:31

moreblue

1738

1738

add a comment | 

add a comment | 

2 Answers
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With observing
$$dfracddalphaleft(dfrac1(k+1)+(n-k-1)e^alpha xright)=dfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2$$
then the desire integral is
beginalign
I
&=dfracalpha-(n-k-1)int_0^inftydfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2dx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrac1(k+1)+(n-k-1)e^alpha xdx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrace^-alpha x(k+1)e^-alpha x+(n-k-1)dx\
&=dfrac1-alpha(k+1)(n-k-1)lndfracn-k-1nendalign

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edited Jul 17 at 13:34

answered Jul 17 at 12:59

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up vote
1
down vote

Well, we have (assuming that the integral exists):

$$mathcalI_spacetextnleft(textk,alpharight):=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(1+textk+expleft(alphacdot xright)cdotleft(textn-textk-1right)right)^2spacetextdxtag1$$

Now, first of all let:

 1. $$beta_1:=1+textktag2$$
 2. $$beta_2:=textn-textk-1tag3$$

So, we can rewrite equation $left(1right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdx=alphacdotint_0^inftyfracxcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdxtag4$$

Now, substitute $textu:=expleft(alphacdot xright)$, so we get:

$$mathcalI_spacetextnleft(textk,alpharight)=alphacdotfrac1alpha^2cdotlim_textptoinftyint_1^expleft(alphacdottextpright)fraclnleft(texturight)left(beta_1+textucdotbeta_2right)^2spacetextdtextutag5$$

Using IBP:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2cdotlefttextZ+lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtexturighttag6$$

Where:

$$textZ:=lim_textptoinftyleft[-fraclnleft(texturight)beta_1+textucdotbeta_2right]_1^expleft(alphacdottextpright)tag7$$

Assuming that $alphainmathbbR$, we can write:

$$textZ=-lim_textptoinftyfracalphacdottextpbeta_1+expleft(alphacdottextpright)cdotbeta_2=0tag8$$

So, we can rewrite equation $left(6right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtextu=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleft[lnleft|fractextubeta_1cdotleft(beta_1+textucdotbeta_2right)right|right]_1^expleft(alphacdottextpright)=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleftfracexpleft(alphacdottextpright)beta_1cdotleft(beta_1+expleft(alphacdottextpright)cdotbeta_2right)right=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftylnleft|fracleft(beta_1+beta_2right)cdotexpleft(alphacdottextpright)beta_1+expleft(alphacdottextpright)cdotbeta_2right|tag9$$

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edited Jul 17 at 13:36

answered Jul 17 at 13:04

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

With observing
$$dfracddalphaleft(dfrac1(k+1)+(n-k-1)e^alpha xright)=dfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2$$
then the desire integral is
beginalign
I
&=dfracalpha-(n-k-1)int_0^inftydfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2dx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrac1(k+1)+(n-k-1)e^alpha xdx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrace^-alpha x(k+1)e^-alpha x+(n-k-1)dx\
&=dfrac1-alpha(k+1)(n-k-1)lndfracn-k-1nendalign

share|cite|improve this answer

edited Jul 17 at 13:34

answered Jul 17 at 12:59

Nosrati

19.8k41644

add a comment | 

up vote
1
down vote

With observing
$$dfracddalphaleft(dfrac1(k+1)+(n-k-1)e^alpha xright)=dfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2$$
then the desire integral is
beginalign
I
&=dfracalpha-(n-k-1)int_0^inftydfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2dx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrac1(k+1)+(n-k-1)e^alpha xdx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrace^-alpha x(k+1)e^-alpha x+(n-k-1)dx\
&=dfrac1-alpha(k+1)(n-k-1)lndfracn-k-1nendalign

share|cite|improve this answer

edited Jul 17 at 13:34

answered Jul 17 at 12:59

Nosrati

19.8k41644

add a comment | 

up vote
1
down vote

up vote
1
down vote

With observing
$$dfracddalphaleft(dfrac1(k+1)+(n-k-1)e^alpha xright)=dfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2$$
then the desire integral is
beginalign
I
&=dfracalpha-(n-k-1)int_0^inftydfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2dx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrac1(k+1)+(n-k-1)e^alpha xdx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrace^-alpha x(k+1)e^-alpha x+(n-k-1)dx\
&=dfrac1-alpha(k+1)(n-k-1)lndfracn-k-1nendalign

share|cite|improve this answer

edited Jul 17 at 13:34

answered Jul 17 at 12:59

Nosrati

19.8k41644

With observing
$$dfracddalphaleft(dfrac1(k+1)+(n-k-1)e^alpha xright)=dfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2$$
then the desire integral is
beginalign
I
&=dfracalpha-(n-k-1)int_0^inftydfrac-(n-k-1)xe^alpha xleft((k+1)+(n-k-1)e^alpha xright)^2dx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrac1(k+1)+(n-k-1)e^alpha xdx\
&=dfracalpha-(n-k-1)dfracddalphaint_0^inftydfrace^-alpha x(k+1)e^-alpha x+(n-k-1)dx\
&=dfrac1-alpha(k+1)(n-k-1)lndfracn-k-1nendalign

share|cite|improve this answer

edited Jul 17 at 13:34

answered Jul 17 at 12:59

Nosrati

19.8k41644

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 13:34

edited Jul 17 at 13:34

edited Jul 17 at 13:34

answered Jul 17 at 12:59

Nosrati

19.8k41644

answered Jul 17 at 12:59

Nosrati

19.8k41644

answered Jul 17 at 12:59

Nosrati

19.8k41644

19.8k41644

add a comment | 

add a comment | 

up vote
1
down vote

Well, we have (assuming that the integral exists):

$$mathcalI_spacetextnleft(textk,alpharight):=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(1+textk+expleft(alphacdot xright)cdotleft(textn-textk-1right)right)^2spacetextdxtag1$$

Now, first of all let:

 1. $$beta_1:=1+textktag2$$
 2. $$beta_2:=textn-textk-1tag3$$

So, we can rewrite equation $left(1right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdx=alphacdotint_0^inftyfracxcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdxtag4$$

Now, substitute $textu:=expleft(alphacdot xright)$, so we get:

$$mathcalI_spacetextnleft(textk,alpharight)=alphacdotfrac1alpha^2cdotlim_textptoinftyint_1^expleft(alphacdottextpright)fraclnleft(texturight)left(beta_1+textucdotbeta_2right)^2spacetextdtextutag5$$

Using IBP:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2cdotlefttextZ+lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtexturighttag6$$

Where:

$$textZ:=lim_textptoinftyleft[-fraclnleft(texturight)beta_1+textucdotbeta_2right]_1^expleft(alphacdottextpright)tag7$$

Assuming that $alphainmathbbR$, we can write:

$$textZ=-lim_textptoinftyfracalphacdottextpbeta_1+expleft(alphacdottextpright)cdotbeta_2=0tag8$$

So, we can rewrite equation $left(6right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtextu=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleft[lnleft|fractextubeta_1cdotleft(beta_1+textucdotbeta_2right)right|right]_1^expleft(alphacdottextpright)=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleftfracexpleft(alphacdottextpright)beta_1cdotleft(beta_1+expleft(alphacdottextpright)cdotbeta_2right)right=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftylnleft|fracleft(beta_1+beta_2right)cdotexpleft(alphacdottextpright)beta_1+expleft(alphacdottextpright)cdotbeta_2right|tag9$$

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edited Jul 17 at 13:36

answered Jul 17 at 13:04

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up vote
1
down vote

Well, we have (assuming that the integral exists):

$$mathcalI_spacetextnleft(textk,alpharight):=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(1+textk+expleft(alphacdot xright)cdotleft(textn-textk-1right)right)^2spacetextdxtag1$$

Now, first of all let:

 1. $$beta_1:=1+textktag2$$
 2. $$beta_2:=textn-textk-1tag3$$

So, we can rewrite equation $left(1right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdx=alphacdotint_0^inftyfracxcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdxtag4$$

Now, substitute $textu:=expleft(alphacdot xright)$, so we get:

$$mathcalI_spacetextnleft(textk,alpharight)=alphacdotfrac1alpha^2cdotlim_textptoinftyint_1^expleft(alphacdottextpright)fraclnleft(texturight)left(beta_1+textucdotbeta_2right)^2spacetextdtextutag5$$

Using IBP:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2cdotlefttextZ+lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtexturighttag6$$

Where:

$$textZ:=lim_textptoinftyleft[-fraclnleft(texturight)beta_1+textucdotbeta_2right]_1^expleft(alphacdottextpright)tag7$$

Assuming that $alphainmathbbR$, we can write:

$$textZ=-lim_textptoinftyfracalphacdottextpbeta_1+expleft(alphacdottextpright)cdotbeta_2=0tag8$$

So, we can rewrite equation $left(6right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtextu=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleft[lnleft|fractextubeta_1cdotleft(beta_1+textucdotbeta_2right)right|right]_1^expleft(alphacdottextpright)=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleftfracexpleft(alphacdottextpright)beta_1cdotleft(beta_1+expleft(alphacdottextpright)cdotbeta_2right)right=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftylnleft|fracleft(beta_1+beta_2right)cdotexpleft(alphacdottextpright)beta_1+expleft(alphacdottextpright)cdotbeta_2right|tag9$$

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edited Jul 17 at 13:36

answered Jul 17 at 13:04

Jan

21.6k31239

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up vote
1
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up vote
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down vote

Well, we have (assuming that the integral exists):

$$mathcalI_spacetextnleft(textk,alpharight):=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(1+textk+expleft(alphacdot xright)cdotleft(textn-textk-1right)right)^2spacetextdxtag1$$

Now, first of all let:

 1. $$beta_1:=1+textktag2$$
 2. $$beta_2:=textn-textk-1tag3$$

So, we can rewrite equation $left(1right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdx=alphacdotint_0^inftyfracxcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdxtag4$$

Now, substitute $textu:=expleft(alphacdot xright)$, so we get:

$$mathcalI_spacetextnleft(textk,alpharight)=alphacdotfrac1alpha^2cdotlim_textptoinftyint_1^expleft(alphacdottextpright)fraclnleft(texturight)left(beta_1+textucdotbeta_2right)^2spacetextdtextutag5$$

Using IBP:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2cdotlefttextZ+lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtexturighttag6$$

Where:

$$textZ:=lim_textptoinftyleft[-fraclnleft(texturight)beta_1+textucdotbeta_2right]_1^expleft(alphacdottextpright)tag7$$

Assuming that $alphainmathbbR$, we can write:

$$textZ=-lim_textptoinftyfracalphacdottextpbeta_1+expleft(alphacdottextpright)cdotbeta_2=0tag8$$

So, we can rewrite equation $left(6right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtextu=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleft[lnleft|fractextubeta_1cdotleft(beta_1+textucdotbeta_2right)right|right]_1^expleft(alphacdottextpright)=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleftfracexpleft(alphacdottextpright)beta_1cdotleft(beta_1+expleft(alphacdottextpright)cdotbeta_2right)right=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftylnleft|fracleft(beta_1+beta_2right)cdotexpleft(alphacdottextpright)beta_1+expleft(alphacdottextpright)cdotbeta_2right|tag9$$

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edited Jul 17 at 13:36

answered Jul 17 at 13:04

Jan

21.6k31239

Well, we have (assuming that the integral exists):

$$mathcalI_spacetextnleft(textk,alpharight):=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(1+textk+expleft(alphacdot xright)cdotleft(textn-textk-1right)right)^2spacetextdxtag1$$

Now, first of all let:

 1. $$beta_1:=1+textktag2$$
 2. $$beta_2:=textn-textk-1tag3$$

So, we can rewrite equation $left(1right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=int_0^inftyfracalphacdot xcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdx=alphacdotint_0^inftyfracxcdotexpleft(alphacdot xright)left(beta_1+expleft(alphacdot xright)cdotbeta_2right)^2spacetextdxtag4$$

Now, substitute $textu:=expleft(alphacdot xright)$, so we get:

$$mathcalI_spacetextnleft(textk,alpharight)=alphacdotfrac1alpha^2cdotlim_textptoinftyint_1^expleft(alphacdottextpright)fraclnleft(texturight)left(beta_1+textucdotbeta_2right)^2spacetextdtextutag5$$

Using IBP:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2cdotlefttextZ+lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtexturighttag6$$

Where:

$$textZ:=lim_textptoinftyleft[-fraclnleft(texturight)beta_1+textucdotbeta_2right]_1^expleft(alphacdottextpright)tag7$$

Assuming that $alphainmathbbR$, we can write:

$$textZ=-lim_textptoinftyfracalphacdottextpbeta_1+expleft(alphacdottextpright)cdotbeta_2=0tag8$$

So, we can rewrite equation $left(6right)$ as follows:

$$mathcalI_spacetextnleft(textk,alpharight)=frac1alphacdotbeta_2lim_textptoinftyint_1^expleft(alphacdottextpright)frac1textucdotleft(beta_1+textucdotbeta_2right)spacetextdtextu=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleft[lnleft|fractextubeta_1cdotleft(beta_1+textucdotbeta_2right)right|right]_1^expleft(alphacdottextpright)=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftyleftfracexpleft(alphacdottextpright)beta_1cdotleft(beta_1+expleft(alphacdottextpright)cdotbeta_2right)right=$$
$$frac1alphacdotbeta_1cdotbeta_2cdotlim_textptoinftylnleft|fracleft(beta_1+beta_2right)cdotexpleft(alphacdottextpright)beta_1+expleft(alphacdottextpright)cdotbeta_2right|tag9$$

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edited Jul 17 at 13:36

answered Jul 17 at 13:04

Jan

21.6k31239

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 13:36

edited Jul 17 at 13:36

edited Jul 17 at 13:36

answered Jul 17 at 13:04

Jan

21.6k31239

answered Jul 17 at 13:04

Jan

21.6k31239

answered Jul 17 at 13:04

Jan

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21.6k31239

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