Frullani integral for $f(x)=e^x$ in a complex contextFrullani integral for $f(x)=e^-lambda x $Calculating integral using cauchy integral formulaAnalytically continuing a function defined by an integral to a larger regionfundamental complex integral( in Conway's book)Multivariate Complex Gaussian IntegralComplex path IntegralRiemann Zeta Function integralIntegral of complex logarithm on a disk in the planeMethods for integral calculationRaabe's integral for complex argumentDefininition of the complex integral $int_z_0^z_1f(z) , dz$.
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Frullani integral for $f(x)=e^x$ in a complex context
Frullani integral for $f(x)=e^-lambda x $Calculating integral using cauchy integral formulaAnalytically continuing a function defined by an integral to a larger regionfundamental complex integral( in Conway's book)Multivariate Complex Gaussian IntegralComplex path IntegralRiemann Zeta Function integralIntegral of complex logarithm on a disk in the planeMethods for integral calculationRaabe's integral for complex argumentDefininition of the complex integral $int_z_0^z_1f(z) , dz$.
$begingroup$
I should prove the Frullani integral equality
$$
int_0^infty (1-e^zx) fracbetax e^-gamma xdx = beta log left(1- fraczgammaright)
$$
for $z in mathbbC$ with non-positive real part.
I should first consider $z leq 0$ and use
$$
frace^-gamma x - e^-(gamma - z)xx=int_gamma^gamma - ze^-y xdy$$
and then change the order of integration. These steps are clear (see also Frullani integral for $f(x)=e^-lambda x $).
But then I should use analytic extension to show that the formula is valid for $z in mathbbC$ with non-positive real part. I need help for this step.
Thank you in advance!
calculus integration complex-analysis stochastic-analysis
edited Mar 12 at 11:46
mathreadler
15.1k72263
asked Mar 12 at 8:54
ChristinaChristina
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I should prove the Frullani integral equality
$$
int_0^infty (1-e^zx) fracbetax e^-gamma xdx = beta log left(1- fraczgammaright)
$$
for $z in mathbbC$ with non-positive real part.
I should first consider $z leq 0$ and use
$$
frace^-gamma x - e^-(gamma - z)xx=int_gamma^gamma - ze^-y xdy$$
and then change the order of integration. These steps are clear (see also Frullani integral for $f(x)=e^-lambda x $).
But then I should use analytic extension to show that the formula is valid for $z in mathbbC$ with non-positive real part. I need help for this step.
Thank you in advance!
calculus integration complex-analysis stochastic-analysis
edited Mar 12 at 11:46
mathreadler
15.1k72263
asked Mar 12 at 8:54
ChristinaChristina
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Should that be $dx$ in your second equation?
$endgroup$
– Bladewood
Mar 12 at 9:22
1
$begingroup$
No, the $dy$ was correct, but I did another mistake: it is not $e^-gamma x$ in the integral, but $e^-yx$, thanks for pointing that out! I will edit it.
$endgroup$
– Christina
Mar 12 at 10:24
add a comment |
$begingroup$
I should prove the Frullani integral equality
$$
int_0^infty (1-e^zx) fracbetax e^-gamma xdx = beta log left(1- fraczgammaright)
$$
for $z in mathbbC$ with non-positive real part.
I should first consider $z leq 0$ and use
$$
frace^-gamma x - e^-(gamma - z)xx=int_gamma^gamma - ze^-y xdy$$
and then change the order of integration. These steps are clear (see also Frullani integral for $f(x)=e^-lambda x $).
But then I should use analytic extension to show that the formula is valid for $z in mathbbC$ with non-positive real part. I need help for this step.
Thank you in advance!
calculus integration complex-analysis stochastic-analysis
edited Mar 12 at 11:46
mathreadler
15.1k72263
asked Mar 12 at 8:54
ChristinaChristina
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I should prove the Frullani integral equality
$$
int_0^infty (1-e^zx) fracbetax e^-gamma xdx = beta log left(1- fraczgammaright)
$$
for $z in mathbbC$ with non-positive real part.
I should first consider $z leq 0$ and use
$$
frace^-gamma x - e^-(gamma - z)xx=int_gamma^gamma - ze^-y xdy$$
and then change the order of integration. These steps are clear (see also Frullani integral for $f(x)=e^-lambda x $).
But then I should use analytic extension to show that the formula is valid for $z in mathbbC$ with non-positive real part. I need help for this step.
Thank you in advance!
calculus integration complex-analysis stochastic-analysis
calculus integration complex-analysis stochastic-analysis
edited Mar 12 at 11:46
mathreadler
15.1k72263
asked Mar 12 at 8:54
ChristinaChristina
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 12 at 11:46
mathreadler
15.1k72263
asked Mar 12 at 8:54
ChristinaChristina
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 12 at 11:46
mathreadler
15.1k72263
edited Mar 12 at 11:46
mathreadler
15.1k72263
edited Mar 12 at 11:46
mathreadler
15.1k72263
15.1k72263
asked Mar 12 at 8:54
ChristinaChristina
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 12 at 8:54
ChristinaChristina
234
asked Mar 12 at 8:54
ChristinaChristina
234
234
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Christina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Should that be $dx$ in your second equation?
$endgroup$
– Bladewood
Mar 12 at 9:22
1
$begingroup$
No, the $dy$ was correct, but I did another mistake: it is not $e^-gamma x$ in the integral, but $e^-yx$, thanks for pointing that out! I will edit it.
$endgroup$
– Christina
Mar 12 at 10:24
add a comment |
2
$begingroup$
Should that be $dx$ in your second equation?
$endgroup$
– Bladewood
Mar 12 at 9:22
1
$begingroup$
No, the $dy$ was correct, but I did another mistake: it is not $e^-gamma x$ in the integral, but $e^-yx$, thanks for pointing that out! I will edit it.
$endgroup$
– Christina
Mar 12 at 10:24
2
2
$begingroup$
Should that be $dx$ in your second equation?
$endgroup$
– Bladewood
Mar 12 at 9:22
$begingroup$
Should that be $dx$ in your second equation?
$endgroup$
– Bladewood
Mar 12 at 9:22
1
1
$begingroup$
No, the $dy$ was correct, but I did another mistake: it is not $e^-gamma x$ in the integral, but $e^-yx$, thanks for pointing that out! I will edit it.
$endgroup$
– Christina
Mar 12 at 10:24
$begingroup$
No, the $dy$ was correct, but I did another mistake: it is not $e^-gamma x$ in the integral, but $e^-yx$, thanks for pointing that out! I will edit it.
$endgroup$
– Christina
Mar 12 at 10:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag1$$
For $xto0$ the integrand remains finite, for $xtoinfty$ convergence requires that
$$Re(gamma)gt0tag2a$$ and $$Re(gamma - z)gt0tag2b$$.
Now the derivative with respect to $gamma$ is
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dxtag3$$
Notice that the $x$ in the denominator has vanished, and the resulting integral is elementary
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dx=-frac1gamma +frac1gamma -ztag4$$
Now integrating with respect to $gamma $ gives
$$ f(gamma,z)= -log(gamma) + log(gamma - z) + C= logleft(fracgamma-zgammaright)=logleft(1-fraczgammaright) tag5$$
The constant $C$ was found to vanish considering the limit $gamma to infty$.
Hence we have proven (apart from the trivial factor $beta$) the relation of the OP under the conditions $(2)$.
$(5)$ then gives the analytic continuation to the complex $z$-plane.
edited Mar 12 at 11:48
mathreadler
15.1k72263
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
1
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
add a comment |
$begingroup$
Using the power series expansion
$$frac(1-e^zx)x = - sum_n=1^infty frac1n! z^n x^n-1tag1$$
the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag2$$
becomes after interchanging sum and integral
$$f(gamma,z)=- sum_n=1^infty frac1n! z^n int_0^infty x^n-1 e^-gamma x,dxtag3$$
Now
$$int_0^infty x^n-1 e^-gamma x,dx = frac1gamma^n Gamma(n)= frac1gamma^n (n-1)!tag4 $$
so that
$$f(gamma,z)=- sum_n=1^infty frac1n!(n-1)! left(fraczgammaright)^n = - sum_n=1^infty frac1n left(fraczgammaright)^n= logleft(1-fraczgammaright)tag5$$
QED.
edited Mar 12 at 20:17
Bladewood
337213
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag1$$
For $xto0$ the integrand remains finite, for $xtoinfty$ convergence requires that
$$Re(gamma)gt0tag2a$$ and $$Re(gamma - z)gt0tag2b$$.
Now the derivative with respect to $gamma$ is
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dxtag3$$
Notice that the $x$ in the denominator has vanished, and the resulting integral is elementary
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dx=-frac1gamma +frac1gamma -ztag4$$
Now integrating with respect to $gamma $ gives
$$ f(gamma,z)= -log(gamma) + log(gamma - z) + C= logleft(fracgamma-zgammaright)=logleft(1-fraczgammaright) tag5$$
The constant $C$ was found to vanish considering the limit $gamma to infty$.
Hence we have proven (apart from the trivial factor $beta$) the relation of the OP under the conditions $(2)$.
$(5)$ then gives the analytic continuation to the complex $z$-plane.
edited Mar 12 at 11:48
mathreadler
15.1k72263
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
1
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
add a comment |
$begingroup$
Consider the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag1$$
For $xto0$ the integrand remains finite, for $xtoinfty$ convergence requires that
$$Re(gamma)gt0tag2a$$ and $$Re(gamma - z)gt0tag2b$$.
Now the derivative with respect to $gamma$ is
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dxtag3$$
Notice that the $x$ in the denominator has vanished, and the resulting integral is elementary
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dx=-frac1gamma +frac1gamma -ztag4$$
Now integrating with respect to $gamma $ gives
$$ f(gamma,z)= -log(gamma) + log(gamma - z) + C= logleft(fracgamma-zgammaright)=logleft(1-fraczgammaright) tag5$$
The constant $C$ was found to vanish considering the limit $gamma to infty$.
Hence we have proven (apart from the trivial factor $beta$) the relation of the OP under the conditions $(2)$.
$(5)$ then gives the analytic continuation to the complex $z$-plane.
edited Mar 12 at 11:48
mathreadler
15.1k72263
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
1
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
add a comment |
$begingroup$
Consider the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag1$$
For $xto0$ the integrand remains finite, for $xtoinfty$ convergence requires that
$$Re(gamma)gt0tag2a$$ and $$Re(gamma - z)gt0tag2b$$.
Now the derivative with respect to $gamma$ is
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dxtag3$$
Notice that the $x$ in the denominator has vanished, and the resulting integral is elementary
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dx=-frac1gamma +frac1gamma -ztag4$$
Now integrating with respect to $gamma $ gives
$$ f(gamma,z)= -log(gamma) + log(gamma - z) + C= logleft(fracgamma-zgammaright)=logleft(1-fraczgammaright) tag5$$
The constant $C$ was found to vanish considering the limit $gamma to infty$.
Hence we have proven (apart from the trivial factor $beta$) the relation of the OP under the conditions $(2)$.
$(5)$ then gives the analytic continuation to the complex $z$-plane.
edited Mar 12 at 11:48
mathreadler
15.1k72263
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
Consider the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag1$$
For $xto0$ the integrand remains finite, for $xtoinfty$ convergence requires that
$$Re(gamma)gt0tag2a$$ and $$Re(gamma - z)gt0tag2b$$.
Now the derivative with respect to $gamma$ is
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dxtag3$$
Notice that the $x$ in the denominator has vanished, and the resulting integral is elementary
$$fracpartial f(gamma,z)partialgamma=-int_0^infty (1-e^zx) e^-gamma x,dx=-frac1gamma +frac1gamma -ztag4$$
Now integrating with respect to $gamma $ gives
$$ f(gamma,z)= -log(gamma) + log(gamma - z) + C= logleft(fracgamma-zgammaright)=logleft(1-fraczgammaright) tag5$$
The constant $C$ was found to vanish considering the limit $gamma to infty$.
Hence we have proven (apart from the trivial factor $beta$) the relation of the OP under the conditions $(2)$.
$(5)$ then gives the analytic continuation to the complex $z$-plane.
edited Mar 12 at 11:48
mathreadler
15.1k72263
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
edited Mar 12 at 11:48
mathreadler
15.1k72263
edited Mar 12 at 11:48
mathreadler
15.1k72263
edited Mar 12 at 11:48
mathreadler
15.1k72263
15.1k72263
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
answered Mar 12 at 10:13
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
3,735620
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
1
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
add a comment |
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
1
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
$begingroup$
thanks for your answer! Can you explain a little bit further why the analytic continuation to the complex-z-plane follows from (5)?
$endgroup$
– Christina
Mar 13 at 9:51
1
1
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
$begingroup$
@ Christina Because (for constant $gamma$ >0 for definiteness) the RHS is an analytic function of $z$. The only singularity is a (logarithmic) branch point at $z=gamma$ with a branch cut from there to $+infty$. Notice that the original integral $(1)$ is convergent only if $(2b)$ holds. Hence $(5)$ is the analytic extension of $(1)$. You can go to the region $Re(z)gt gamma$ above or below the branch cut.
$endgroup$
– Dr. Wolfgang Hintze
2 days ago
add a comment |
$begingroup$
Using the power series expansion
$$frac(1-e^zx)x = - sum_n=1^infty frac1n! z^n x^n-1tag1$$
the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag2$$
becomes after interchanging sum and integral
$$f(gamma,z)=- sum_n=1^infty frac1n! z^n int_0^infty x^n-1 e^-gamma x,dxtag3$$
Now
$$int_0^infty x^n-1 e^-gamma x,dx = frac1gamma^n Gamma(n)= frac1gamma^n (n-1)!tag4 $$
so that
$$f(gamma,z)=- sum_n=1^infty frac1n!(n-1)! left(fraczgammaright)^n = - sum_n=1^infty frac1n left(fraczgammaright)^n= logleft(1-fraczgammaright)tag5$$
QED.
edited Mar 12 at 20:17
Bladewood
337213
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
add a comment |
$begingroup$
Using the power series expansion
$$frac(1-e^zx)x = - sum_n=1^infty frac1n! z^n x^n-1tag1$$
the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag2$$
becomes after interchanging sum and integral
$$f(gamma,z)=- sum_n=1^infty frac1n! z^n int_0^infty x^n-1 e^-gamma x,dxtag3$$
Now
$$int_0^infty x^n-1 e^-gamma x,dx = frac1gamma^n Gamma(n)= frac1gamma^n (n-1)!tag4 $$
so that
$$f(gamma,z)=- sum_n=1^infty frac1n!(n-1)! left(fraczgammaright)^n = - sum_n=1^infty frac1n left(fraczgammaright)^n= logleft(1-fraczgammaright)tag5$$
QED.
edited Mar 12 at 20:17
Bladewood
337213
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
add a comment |
$begingroup$
Using the power series expansion
$$frac(1-e^zx)x = - sum_n=1^infty frac1n! z^n x^n-1tag1$$
the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag2$$
becomes after interchanging sum and integral
$$f(gamma,z)=- sum_n=1^infty frac1n! z^n int_0^infty x^n-1 e^-gamma x,dxtag3$$
Now
$$int_0^infty x^n-1 e^-gamma x,dx = frac1gamma^n Gamma(n)= frac1gamma^n (n-1)!tag4 $$
so that
$$f(gamma,z)=- sum_n=1^infty frac1n!(n-1)! left(fraczgammaright)^n = - sum_n=1^infty frac1n left(fraczgammaright)^n= logleft(1-fraczgammaright)tag5$$
QED.
edited Mar 12 at 20:17
Bladewood
337213
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
$endgroup$
Using the power series expansion
$$frac(1-e^zx)x = - sum_n=1^infty frac1n! z^n x^n-1tag1$$
the integral
$$f(gamma,z)=int_0^infty frac(1-e^zx)x e^-gamma x,dxtag2$$
becomes after interchanging sum and integral
$$f(gamma,z)=- sum_n=1^infty frac1n! z^n int_0^infty x^n-1 e^-gamma x,dxtag3$$
Now
$$int_0^infty x^n-1 e^-gamma x,dx = frac1gamma^n Gamma(n)= frac1gamma^n (n-1)!tag4 $$
so that
$$f(gamma,z)=- sum_n=1^infty frac1n!(n-1)! left(fraczgammaright)^n = - sum_n=1^infty frac1n left(fraczgammaright)^n= logleft(1-fraczgammaright)tag5$$
QED.
edited Mar 12 at 20:17
Bladewood
337213
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
edited Mar 12 at 20:17
Bladewood
337213
edited Mar 12 at 20:17
Bladewood
337213
edited Mar 12 at 20:17
Bladewood
337213
337213
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
answered Mar 12 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,735620
3,735620
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Christina is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
Should that be $dx$ in your second equation?
$endgroup$
– Bladewood
Mar 12 at 9:22
1
$begingroup$
No, the $dy$ was correct, but I did another mistake: it is not $e^-gamma x$ in the integral, but $e^-yx$, thanks for pointing that out! I will edit it.
$endgroup$
– Christina
Mar 12 at 10:24