The equation Γ ′ ( x ) = ∫ 0 ∞ t x − 1 e − t ln t d t and Γ ″ ( x ) = ∫ 0 ∞ t x − 1 e − t ( ln t ) 2 d t by the use of gamma function and Leibniz’s rule for x > 0 .

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Answer 1

Question

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

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Answer 2

Question

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

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Answer 3

Question

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

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Answer 4

Question

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

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Answer 5

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

Answer 6

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

Answer 7

Chapter A, Problem 27E

(a)

To determine

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

Answer 8

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

Answer 9

(b)

To determine

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

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Answer 13

Ch. A - In Problems 1 and 2 evaluate the given quantity....Ch. A - Prob. 2E Ch. A - Prob. 3E Ch. A - Prob. 4E Ch. A - Prob. 5E Ch. A - Prob. 6E Ch. A - Prob. 7E Ch. A - Prob. 8E Ch. A - Prob. 9E Ch. A - Prob. 10E

The equation Γ ′ ( x ) = ∫ 0 ∞ t x − 1 e − t ln t d t and Γ ″ ( x ) = ∫ 0 ∞ t x − 1 e − t ( ln t ) 2 d t by the use of gamma function and Leibniz’s rule for x > 0 .

Solution Summary: The author explains the gamma function and Leibniz's rule for x>0.

Concept explainers

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To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

Want to see the full answer?

Chapter A Solutions

The equation Γ ′ ( x ) = ∫ 0 ∞ t x − 1 e − t ln t d t and Γ ″ ( x ) = ∫ 0 ∞ t x − 1 e − t ( ln t ) 2 d t by the use of gamma function and Leibniz’s rule for x &gt; 0 .

Solution Summary: The author explains the gamma function and Leibniz's rule for x>0.

Concept explainers

Videos

To show: The equation Γ′(x)=∫0∞tx−1e−tlntdt and Γ″(x)=∫0∞tx−1e−t(lnt)2dt by the use of gamma function and Leibniz’s rule for x>0.

The behavior of the graph of a Γ(x), if for x>0, Γ(x)>0 and Γ″(x)>0.

Want to see the full answer?

Chapter A Solutions

The equation Γ ′ ( x ) = ∫ 0 ∞ t x − 1 e − t ln t d t and Γ ″ ( x ) = ∫ 0 ∞ t x − 1 e − t ( ln t ) 2 d t by the use of gamma function and Leibniz’s rule for x > 0 .