How would I go about evaluating this integral? ∫0∞ln⁡(x2+1)x2+1dx

Question

How would I go about evaluating this integral?  ∫0∞ln⁡(x2+1)x2+1dx

Anzante2m · Accepted Answer

Use x→tan⁡θ and dx=sec2θdθ. The integral becomes  ∫0π22ln⁡(sec⁡θ)dθ  Which is  ∫0π2ln⁡(cos⁡θ)dθ  And can be solved  Let  I=∫0π2ln⁡(sin⁡θ)⋅dθ  I=∫0π2ln⁡(cos⁡θ)⋅dθ  Adding both.  2I=∫0π2ln⁡(sin⁡θ×cos⁡θ)⋅dθ  2I=∫0π2ln⁡(2sin⁡θ×cos⁡θ)−ln⁡2⋅dθ  2I=∫0π2ln⁡(sin⁡2θ)⋅dθ=I  So,  I=−∫0π2ln⁡2⋅dθ  I=−πln⁡22  And your integral comes out to be  πln⁡2

Lakisha Archer · Accepted Answer

One way to solve this problem is to use parametric integrals. LetI(α)=∫0∞ln⁡(αx2+1)x2+1dxThenI′(α)=∫0∞x2(αx2+1)(x2+1)dx=∫0∞(−1α−11αx2+1+1α−11x2+1)dx=−1α−11απ2+1α−1π2=π2α−1α(α−1)=π2(1α−1α+1)Thus,I(α)=πln⁡(α+1)+CBut I(0)=0 implies C=0. So I(1)=πln⁡2

Vasquez · Accepted Answer

Hints As RandomVariable suggested, use log⁡(x2+1)=log⁡(x+i)+log⁡(x−i), choosing the branches of the logarithm carefully. It&#039;s generally best to isolate unpleasant singular things. Then write ∫0∞=12∫−∞∞ and make use of the above splitting to integrate the two parts on different contours, each time avoiding enclosing the singularity of the logarithm. The part on the semicircle vanishes. Answer The UHP pole gives log⁡(i+i)/2i×2πi. The LHP pole gives log⁡(−i−i)/(−2i)×−2πi. Summing and halving gives the answer π2(log⁡(2i)+log⁡(−2i)) so all that remains is choosing the right logarithm. This is easy enough, actually, and the answer is what you expect: πln⁡2

How would I go about evaluating this integral? \int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx

Answered question

Answer & Explanation

New Questions in Calculus 1