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Ive been struggling with this for days. Guys, enlighten me!

Note by Danny Kills 2 years, 10 months ago

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Let

\(I(a)=\displaystyle\int_{0}^{\infty} \frac{\ln (1+a^{2}x^{2})}{1+b^{2}x^{2}}\text{ }\text{d}x\)

Differentiating w.r.t. \(a\):

\(I'(a)=\displaystyle\int_{0}^{\infty} \frac{2ax^{2}}{(1+a^{2}x^{2})(1+b^{2}x^{2})}\text{ }\text{d}x\)

\(\therefore I'(a)=\displaystyle\frac{\pi}{b(b+a)}\)

\(\therefore I(a)=\dfrac{\pi}{b}\ln (a+b)+\mathcal{C}\)

Substituting \(a=0\) we obtain the value of \(\mathcal{C}\) as \(-\dfrac{\pi}{b}\ln b\). – Karthik Kannan · 2 years, 9 months ago

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@Karthik Kannan – A really elegant solution. Thanks! – Danny Kills · 2 years, 9 months ago

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TopNewestLet

\(I(a)=\displaystyle\int_{0}^{\infty} \frac{\ln (1+a^{2}x^{2})}{1+b^{2}x^{2}}\text{ }\text{d}x\)

Differentiating w.r.t. \(a\):

\(I'(a)=\displaystyle\int_{0}^{\infty} \frac{2ax^{2}}{(1+a^{2}x^{2})(1+b^{2}x^{2})}\text{ }\text{d}x\)

\(\therefore I'(a)=\displaystyle\frac{\pi}{b(b+a)}\)

\(\therefore I(a)=\dfrac{\pi}{b}\ln (a+b)+\mathcal{C}\)

Substituting \(a=0\) we obtain the value of \(\mathcal{C}\) as \(-\dfrac{\pi}{b}\ln b\). – Karthik Kannan · 2 years, 9 months ago

Log in to reply

– Danny Kills · 2 years, 9 months ago

A really elegant solution. Thanks!Log in to reply