Question #37ae3

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Answer 1 · 2017-10-24T03:42:02

$0 .$ $0.$

Suppose that, $I = \int_{0}^{\infty} \frac{ln x}{1 + x^{2}} d x .$ $I=int_0^oolnx/(1+x^2)dx.$

We subst. $x=tany.:. dx=sec^2ydy.$

Further, $x=0 rArr y=0,"&, as "x to oo, y to pi/2.$

$:. I=int_0^(pi/2) (lntany/cancel(1+tan^2y))*cancel(sec^2y)dy, i.e.,$

$I=int_0^(pi/2)lntanydy.........(star^1).$

Now, we know that, $int_0^af(y)dx=int_0^af(a-y)dx....(R).$

We apply the Result R to $(star^1)"with "a=pi/2, &, f(y)=lntany.$

$:. I=int_0^(pi/2)lntan(pi/2-y)dy, or,$

$I=int_0^(pi/2)lncotydy...................(star^2).$

$:. (star^1)+(star^2) rArr 2I=int_0^(pi/2)lntanydy+int_0^(pi/2)lncotydy,$

$:. 2I=int_0^(pi/2)[lntany+lncoty}dy,$

$=int_0^(pi/2)[ln(tany*coty)]dy,$

$=int_0^(pi/2)ln1dy,$

$:. 2I=int_0^(pi/2)0dy=0.$

$rArr I=0.$

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