"How to evaluate: ∫−∞∞ln⁡(x2+1)x2+1dx" was answered by students like you

Lennie Davis

Answered question

2021-12-30

How to evaluate:

\int_{- \infty}^{\infty} \frac{\ln (x^{2} + 1)}{x^{2} + 1} d x

maul124uk

Beginner2021-12-31Added 35 answers

Let

I (α) = \int_{- \infty}^{\infty} \frac{\ln (a x^{2} + 1)}{x^{2} + 1} d x

Then

I^{'} (α) = \int_{- \infty}^{\infty} \frac{x^{2}}{(a x^{2} + 1) (x^{2} + 1)} d x

= \frac{1}{α - 1} \int_{- \infty}^{\infty} [\frac{1}{x^{2} + 1} - \frac{1}{α x^{2} + 1}] d x

= \frac{π}{\sqrt{α} + α}

and hence

I (α) = \int \frac{π}{\sqrt{α} + α} d α

= 2 π \ln (1 + \sqrt{α}) + C

Clearly

I (0) = 0

implies

C = 0

. Thus

\int_{- \infty}^{\infty} \frac{\ln (x^{2} + 1)}{x^{2} + 1} d x = I (1) = 2 π \ln 2

encolatgehu

Beginner2022-01-01Added 27 answers

Letting

x = \tan t

leads to

- 4 \int_{0}^{\frac{π}{2}} \log (\cos t) d t = 2 π \log 2

we could consider:

I (s) = \int_{R} {(1 + x^{2})}^{s} d x

, and then evaluate

I^{'} (- 1)

I (s) = \int_{R} {(1 + x^{2})}^{s} d x = 2 \int_{0}^{\infty} {(1 + x^{2})}^{s} d x = \int_{0}^{1} {(1 - u)}^{- \frac{3}{2} - s} \frac{d u}{\sqrt{u}}

= B (\frac{1}{2}, - \frac{1}{2} - s)

Thus we established

I (s) = \sqrt{π} \frac{Γ (- \frac{1}{2} - s)}{Γ (- s)}

. We are now ready to compute the derivative:

I^{'} (s) = I (s) (ψ (- s) - ψ (- s - \frac{1}{2}))

and

I^{'} (- 1) = I (- 1) (ψ (1) - ψ (\frac{1}{2}))

= \sqrt{π} \frac{Γ (\frac{1}{2})}{Γ (1)} (ψ (1) - ψ (\frac{1}{2})) = 2 π \log (2)

where

Γ (\frac{1}{2}) = \sqrt{π}

was used, as well as a polygamma duplication identity:

ψ (2 s) = \log (2) + \frac{1}{2} (ψ (s) + ψ (s + \frac{1}{2}))

That evaluated at

s = \frac{1}{2}

gives

ψ (1) - ψ (\frac{1}{2}) = 2 \log (2)

Vasquez

Expert2022-01-09Added 669 answers

Thanks a lot.

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