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dtiwari
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Pe: 2018-05-05T14:05:22+03:00 In: In:

Definite Integral

\displaystyle \int^{\infty}_{0}\bigg(\frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}\bigg)dx Where a>0

When i solve it \displaystyle I=\int^{\infty}_{0}\frac{\tan^{-1}(ax)}{x}dx-\int^{\infty}_{0}\frac{\tan^{-1}(x)}{x}dx

Put ax=t and adx=dt and changing limits we have

\displaystyle I=\int^{\infty}_{0}\frac{\tan^{-1}(t)}{t}dt-\int^{\infty}_{0}\frac{\tan^{-1}(x)}{x}dx =0

i did not understand whats wrong with my reasoning. please clearfy me . thanks

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3 raspunsuri

  1. PhantomR
    expert (VI)

    This is the problem 🙂: – those integrals do NOT converge.

    • 0
  2. PhantomR
    expert (VI)

    We can consider this function I(a)=\int_0^{\infty} \frac{\arctan(ax)-\arctan x }{x} \ \mathrm{d}x.

    Now, there is some theorem (it may be this: https://en.wikipedia.org/wiki/Leibniz_integral_rule), which allows us to differentiate under the integral sign. I have not checked whether our current integral verifies the theorem’s conditions, but I’m almost sure it does.

    So, differentiating with respect to a, we get: I'(a)=\int_0^\infty \frac{\frac{1}{1+a^2x^2} \cdot x - 0 } { x} \mathrm{d} x=\int_0^{\infty} \frac{1}{a^2x^2} \ \mathrm{d}x =\frac{\arctan(ax)}{a} |^\infty_0= \frac{\pi}{2a}. As a consequence, I(a)=\int \frac{\pi}{2a} \ \mathrm{d}a = \frac{\pi}{2}\ln a + K. Noticing that 0=I(1)=K, we obtain that our initial integral is equal to I(a) = \frac{\pi}{2}\ln a.

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  4. dtiwari
    user (0)

    Thanks PhantomR got it.

    • 0
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