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$int_-1^1 (t-1)left(e^frac1Gamma(t)-1right)dt$

1

$begingroup$

I am looking for a way to evaluate the integral

$$
int_-1^1left(t - 1right)left[mathrme^1/Gammaleft(tright) - 1right]mathrmdt
$$

This integral appears to almost have a sort of symmetry about the $y$-axis that could yield a cancellation, but I have been unable to show this. Numerical integration yields a value of $−0.0001194628623602256$. I am curious if there is a way to evaluate this integral, perhaps exploiting any potential symmetry, but I would not be surprised if there is no such method given the difficulty in working with the gamma function.

I have tried substituting $t=-u$ and breaking up the integral into two parts in order to solve but have so far been unsuccessful.

integration complex-analysis definite-integrals gamma-function

share|cite|improve this question

edited Apr 8 at 21:00

Felix Marin

69k7110147

asked Apr 8 at 20:49

csch2csch2

6391314

$endgroup$

add a comment | 

1

$begingroup$

I am looking for a way to evaluate the integral

$$
int_-1^1left(t - 1right)left[mathrme^1/Gammaleft(tright) - 1right]mathrmdt
$$

This integral appears to almost have a sort of symmetry about the $y$-axis that could yield a cancellation, but I have been unable to show this. Numerical integration yields a value of $−0.0001194628623602256$. I am curious if there is a way to evaluate this integral, perhaps exploiting any potential symmetry, but I would not be surprised if there is no such method given the difficulty in working with the gamma function.

I have tried substituting $t=-u$ and breaking up the integral into two parts in order to solve but have so far been unsuccessful.

integration complex-analysis definite-integrals gamma-function

share|cite|improve this question

edited Apr 8 at 21:00

Felix Marin

69k7110147

asked Apr 8 at 20:49

csch2csch2

6391314

$endgroup$

add a comment | 

$begingroup$

I am looking for a way to evaluate the integral

$$
int_-1^1left(t - 1right)left[mathrme^1/Gammaleft(tright) - 1right]mathrmdt
$$

This integral appears to almost have a sort of symmetry about the $y$-axis that could yield a cancellation, but I have been unable to show this. Numerical integration yields a value of $−0.0001194628623602256$. I am curious if there is a way to evaluate this integral, perhaps exploiting any potential symmetry, but I would not be surprised if there is no such method given the difficulty in working with the gamma function.

I have tried substituting $t=-u$ and breaking up the integral into two parts in order to solve but have so far been unsuccessful.

integration complex-analysis definite-integrals gamma-function

share|cite|improve this question

edited Apr 8 at 21:00

Felix Marin

69k7110147

asked Apr 8 at 20:49

csch2csch2

6391314

$endgroup$

share|cite|improve this question

edited Apr 8 at 21:00

Felix Marin

69k7110147

asked Apr 8 at 20:49

csch2csch2

6391314

share|cite|improve this question

edited Apr 8 at 21:00

Felix Marin

69k7110147

asked Apr 8 at 20:49

csch2csch2

6391314

edited Apr 8 at 21:00

Felix Marin

69k7110147

edited Apr 8 at 21:00

Felix Marin

69k7110147

asked Apr 8 at 20:49

csch2csch2

6391314

asked Apr 8 at 20:49

csch2csch2

6391314

1 Answer
1

active

oldest

votes

1

$begingroup$

You can get this kind of small numbers using the Taylor expansion of $e^frac1Gamma(t)$ built at $t=0$ and integrate termwise. This would be quite long; the first terms are
$$e^frac1Gamma(t)-1=t+left(frac12+gamma right) t^2+left(frac16+gamma +fracgamma
^22-fracpi ^212right) t^3+$$ $$frac124 left(1+12 gamma +24
gamma ^2+4 gamma ^3-2 pi ^2-2 gamma pi ^2-4 psi
^(2)(1)right)t^4+Oleft(t^5right)$$

Let us admit that you are sufficiently patient to do the expansion up to $Oleft(t^p+1right)$. You should get
$$left(
beginarraycc
p & textresult \
2 & -0.05147711 \
4 & +0.06413149 \
6 & -0.00956415 \
8 & -0.00483854 \
10 & +0.00118575 \
12 & -0.00041961 \
14 & -0.00003656
endarray
right)$$ I give up (my computer too !).

share|cite|improve this answer

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135

$endgroup$

add a comment | 

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1

$begingroup$

You can get this kind of small numbers using the Taylor expansion of $e^frac1Gamma(t)$ built at $t=0$ and integrate termwise. This would be quite long; the first terms are
$$e^frac1Gamma(t)-1=t+left(frac12+gamma right) t^2+left(frac16+gamma +fracgamma
^22-fracpi ^212right) t^3+$$ $$frac124 left(1+12 gamma +24
gamma ^2+4 gamma ^3-2 pi ^2-2 gamma pi ^2-4 psi
^(2)(1)right)t^4+Oleft(t^5right)$$

Let us admit that you are sufficiently patient to do the expansion up to $Oleft(t^p+1right)$. You should get
$$left(
beginarraycc
p & textresult \
2 & -0.05147711 \
4 & +0.06413149 \
6 & -0.00956415 \
8 & -0.00483854 \
10 & +0.00118575 \
12 & -0.00041961 \
14 & -0.00003656
endarray
right)$$ I give up (my computer too !).

share|cite|improve this answer

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135

$endgroup$

add a comment | 

1

$begingroup$

You can get this kind of small numbers using the Taylor expansion of $e^frac1Gamma(t)$ built at $t=0$ and integrate termwise. This would be quite long; the first terms are
$$e^frac1Gamma(t)-1=t+left(frac12+gamma right) t^2+left(frac16+gamma +fracgamma
^22-fracpi ^212right) t^3+$$ $$frac124 left(1+12 gamma +24
gamma ^2+4 gamma ^3-2 pi ^2-2 gamma pi ^2-4 psi
^(2)(1)right)t^4+Oleft(t^5right)$$

Let us admit that you are sufficiently patient to do the expansion up to $Oleft(t^p+1right)$. You should get
$$left(
beginarraycc
p & textresult \
2 & -0.05147711 \
4 & +0.06413149 \
6 & -0.00956415 \
8 & -0.00483854 \
10 & +0.00118575 \
12 & -0.00041961 \
14 & -0.00003656
endarray
right)$$ I give up (my computer too !).

share|cite|improve this answer

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135

$endgroup$

add a comment | 

$begingroup$

You can get this kind of small numbers using the Taylor expansion of $e^frac1Gamma(t)$ built at $t=0$ and integrate termwise. This would be quite long; the first terms are
$$e^frac1Gamma(t)-1=t+left(frac12+gamma right) t^2+left(frac16+gamma +fracgamma
^22-fracpi ^212right) t^3+$$ $$frac124 left(1+12 gamma +24
gamma ^2+4 gamma ^3-2 pi ^2-2 gamma pi ^2-4 psi
^(2)(1)right)t^4+Oleft(t^5right)$$

Let us admit that you are sufficiently patient to do the expansion up to $Oleft(t^p+1right)$. You should get
$$left(
beginarraycc
p & textresult \
2 & -0.05147711 \
4 & +0.06413149 \
6 & -0.00956415 \
8 & -0.00483854 \
10 & +0.00118575 \
12 & -0.00041961 \
14 & -0.00003656
endarray
right)$$ I give up (my computer too !).

share|cite|improve this answer

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135

$endgroup$

share|cite|improve this answer

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135

answered Apr 9 at 5:52

Claude LeiboviciClaude Leibovici

126k1158135