Evaluating $int_{-a}^a frac{f(x)+1}{b^x+1}$ for a y-axis symmetrical function f

1

$begingroup$

Let $[-a, a] subset mathbb{R}$ be an interval. Let $f: [-a, a] rightarrow mathbb{R}$ be a Riemann-integrable function such that $f(x) = f(-x) space forall x in [-a, a]$. Let $b in mathbb{R}_+$.

Question: Is there a "good" way to simplify the integral $int_{-a}^a frac{f(x) + 1}{b^x+1}$ dx?

calculus integration

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asked Jan 6 at 14:45

Neil WernerNeil Werner

83

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly helpful: What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even function $f$?.
  $endgroup$
  – Martin R
  Jan 6 at 14:47
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $[-a, a] subset mathbb{R}$ be an interval. Let $f: [-a, a] rightarrow mathbb{R}$ be a Riemann-integrable function such that $f(x) = f(-x) space forall x in [-a, a]$. Let $b in mathbb{R}_+$.

Question: Is there a "good" way to simplify the integral $int_{-a}^a frac{f(x) + 1}{b^x+1}$ dx?

calculus integration

share|cite|improve this question

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly helpful: What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even function $f$?.
  $endgroup$
  – Martin R
  Jan 6 at 14:47
  
  
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Let $[-a, a] subset mathbb{R}$ be an interval. Let $f: [-a, a] rightarrow mathbb{R}$ be a Riemann-integrable function such that $f(x) = f(-x) space forall x in [-a, a]$. Let $b in mathbb{R}_+$.

Question: Is there a "good" way to simplify the integral $int_{-a}^a frac{f(x) + 1}{b^x+1}$ dx?

calculus integration

share|cite|improve this question

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

$endgroup$

Let $[-a, a] subset mathbb{R}$ be an interval. Let $f: [-a, a] rightarrow mathbb{R}$ be a Riemann-integrable function such that $f(x) = f(-x) space forall x in [-a, a]$. Let $b in mathbb{R}_+$.

Question: Is there a "good" way to simplify the integral $int_{-a}^a frac{f(x) + 1}{b^x+1}$ dx?

calculus integration

calculus integration

share|cite|improve this question

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

share|cite|improve this question

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

share|cite|improve this question

share|cite|improve this question

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

asked Jan 6 at 14:45

Neil WernerNeil Werner

83

83

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly helpful: What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even function $f$?.
  $endgroup$
  – Martin R
  Jan 6 at 14:47
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly helpful: What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even function $f$?.
  $endgroup$
  – Martin R
  Jan 6 at 14:47
  
  
  

  
  
  
  

$begingroup$
Possibly helpful: What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even function $f$?.
$endgroup$
– Martin R
Jan 6 at 14:47

$begingroup$
Possibly helpful: What functions $g$ satisfy $int_{-L}^{L} frac{f(x)}{1 + g(x)}:dx = int_{0}^{L} f(x):dx$ for every even function $f$?.
$endgroup$
– Martin R
Jan 6 at 14:47

add a comment | 

1 Answer
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3

$begingroup$

Yes! In fact,

$displaystyleint_{-a}^a frac{f(x)+1}{b^x+1} = a+int_0^a f(x) spacemathrm{d}x$

Proof:
We have
begin{equation}label{*}tag{*}
int_{-a}^a frac{f(x)+1}{b^x+1} dx overset{text{substituting } x rightarrow -x}{=} (-1)^2 cdot int_{-a}^a frac{f(x)+1}{b^{-x}+1} spacemathrm{d}x
= int_{-a}^a frac{b^x cdot (f(x) + 1)}{b^x + 1} spacemathrm{d}x
end{equation}

Thus
begin{gather}
displaystyle 2 cdot int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x \ overset{ref{*}}{=} int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x + int_{-a}^a frac{b^x cdot (f(x)+1)}{b^x+1} spacemathrm{d}x \
displaystyle = int_{-a}^a frac{(b^x+1) cdot (f(x)+1)}{b^x+1} spacemathrm{d}x = int_{-a}^a f(x)+1 spacemathrm{d}x = 2 a + int_{-a}^a f(x) spacemathrm{d}x \ text{This achieves a proof}. quad square
end{gather}

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edited Jan 6 at 18:05

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

$begingroup$

Yes! In fact,

$displaystyleint_{-a}^a frac{f(x)+1}{b^x+1} = a+int_0^a f(x) spacemathrm{d}x$

Proof:
We have
begin{equation}label{*}tag{*}
int_{-a}^a frac{f(x)+1}{b^x+1} dx overset{text{substituting } x rightarrow -x}{=} (-1)^2 cdot int_{-a}^a frac{f(x)+1}{b^{-x}+1} spacemathrm{d}x
= int_{-a}^a frac{b^x cdot (f(x) + 1)}{b^x + 1} spacemathrm{d}x
end{equation}

Thus
begin{gather}
displaystyle 2 cdot int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x \ overset{ref{*}}{=} int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x + int_{-a}^a frac{b^x cdot (f(x)+1)}{b^x+1} spacemathrm{d}x \
displaystyle = int_{-a}^a frac{(b^x+1) cdot (f(x)+1)}{b^x+1} spacemathrm{d}x = int_{-a}^a f(x)+1 spacemathrm{d}x = 2 a + int_{-a}^a f(x) spacemathrm{d}x \ text{This achieves a proof}. quad square
end{gather}

share|cite|improve this answer

edited Jan 6 at 18:05

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

$endgroup$

add a comment | 

3

$begingroup$

Yes! In fact,

$displaystyleint_{-a}^a frac{f(x)+1}{b^x+1} = a+int_0^a f(x) spacemathrm{d}x$

Proof:
We have
begin{equation}label{*}tag{*}
int_{-a}^a frac{f(x)+1}{b^x+1} dx overset{text{substituting } x rightarrow -x}{=} (-1)^2 cdot int_{-a}^a frac{f(x)+1}{b^{-x}+1} spacemathrm{d}x
= int_{-a}^a frac{b^x cdot (f(x) + 1)}{b^x + 1} spacemathrm{d}x
end{equation}

Thus
begin{gather}
displaystyle 2 cdot int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x \ overset{ref{*}}{=} int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x + int_{-a}^a frac{b^x cdot (f(x)+1)}{b^x+1} spacemathrm{d}x \
displaystyle = int_{-a}^a frac{(b^x+1) cdot (f(x)+1)}{b^x+1} spacemathrm{d}x = int_{-a}^a f(x)+1 spacemathrm{d}x = 2 a + int_{-a}^a f(x) spacemathrm{d}x \ text{This achieves a proof}. quad square
end{gather}

share|cite|improve this answer

edited Jan 6 at 18:05

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

$endgroup$

add a comment | 

3

3

3

$begingroup$

Yes! In fact,

$displaystyleint_{-a}^a frac{f(x)+1}{b^x+1} = a+int_0^a f(x) spacemathrm{d}x$

Proof:
We have
begin{equation}label{*}tag{*}
int_{-a}^a frac{f(x)+1}{b^x+1} dx overset{text{substituting } x rightarrow -x}{=} (-1)^2 cdot int_{-a}^a frac{f(x)+1}{b^{-x}+1} spacemathrm{d}x
= int_{-a}^a frac{b^x cdot (f(x) + 1)}{b^x + 1} spacemathrm{d}x
end{equation}

Thus
begin{gather}
displaystyle 2 cdot int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x \ overset{ref{*}}{=} int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x + int_{-a}^a frac{b^x cdot (f(x)+1)}{b^x+1} spacemathrm{d}x \
displaystyle = int_{-a}^a frac{(b^x+1) cdot (f(x)+1)}{b^x+1} spacemathrm{d}x = int_{-a}^a f(x)+1 spacemathrm{d}x = 2 a + int_{-a}^a f(x) spacemathrm{d}x \ text{This achieves a proof}. quad square
end{gather}

share|cite|improve this answer

edited Jan 6 at 18:05

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

$endgroup$

Yes! In fact,

$displaystyleint_{-a}^a frac{f(x)+1}{b^x+1} = a+int_0^a f(x) spacemathrm{d}x$

Proof:
We have
begin{equation}label{*}tag{*}
int_{-a}^a frac{f(x)+1}{b^x+1} dx overset{text{substituting } x rightarrow -x}{=} (-1)^2 cdot int_{-a}^a frac{f(x)+1}{b^{-x}+1} spacemathrm{d}x
= int_{-a}^a frac{b^x cdot (f(x) + 1)}{b^x + 1} spacemathrm{d}x
end{equation}

Thus
begin{gather}
displaystyle 2 cdot int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x \ overset{ref{*}}{=} int_{-a}^a frac{f(x)+1}{b^x+1} spacemathrm{d}x + int_{-a}^a frac{b^x cdot (f(x)+1)}{b^x+1} spacemathrm{d}x \
displaystyle = int_{-a}^a frac{(b^x+1) cdot (f(x)+1)}{b^x+1} spacemathrm{d}x = int_{-a}^a f(x)+1 spacemathrm{d}x = 2 a + int_{-a}^a f(x) spacemathrm{d}x \ text{This achieves a proof}. quad square
end{gather}

share|cite|improve this answer

edited Jan 6 at 18:05

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

share|cite|improve this answer

share|cite|improve this answer

edited Jan 6 at 18:05

edited Jan 6 at 18:05

edited Jan 6 at 18:05

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

answered Jan 6 at 14:52

Maximilian JanischMaximilian Janisch

44110

44110

add a comment | 

add a comment | 

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