Question-219550

Question Number 219550 by Spillover last updated on 28/Apr/25

Answered by Nicholas666 last updated on 29/Apr/25

I=∫_0 ^( ∞) ((ln(1+x^7 )ln(1+x^3 ))/((1+x^2 )lnx))dx I(a)=∫_0 ^( ∞) ((ln(1+x^a ))/((1+x^2 )lnx))dx ((dI(a))/da)=∫_0 ^∞ (x^a /((1+x^2 )(1+x^a )))dx ((dI(a))/da)=∫_0 ^∞ (1/((1+x^2 )(1+x^a )))dx 2((dI(a))/da)=∫_0 ^∞ ((x^a +1)/((1+x^2 )(1+x^a )))dx=∫_0 ^∞ (1/(1+x^2 ))dx 2((dI(a))/da)=[arctan x]_0 ^∞ =(π/2) ((dI(a))/da)=(π/4) I(a)=(π/4)a+C I=I(7)−I(3)=((π/4).7+C)−((π/4).3+C) I=((7π)/4)−((3π)/4)=((4π)/4)=π Final Answer ∫_0 ^∞ ((ln(1+x^7 )−ln(1+x^3 ))/((1+x^2 )lnx))dx=π

$${I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{7}} \right){ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){lnx}}{dx}\:\:\:\:\:\:\: \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}^{{a}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){lnx}}{dx} \\ $$$$\frac{{dI}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{{a}} \right)}{dx} \\ $$$$\frac{{dI}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{{a}} \right)}{dx} \\ $$$$\mathrm{2}\frac{{dI}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} +\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{{a}} \right)}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\: \\ $$$$\mathrm{2}\frac{{dI}\left({a}\right)}{{da}}=\left[{arctan}\:{x}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}} \\ $$$$\frac{{dI}\left({a}\right)}{{da}}=\frac{\pi}{\mathrm{4}} \\ $$$${I}\left({a}\right)=\frac{\pi}{\mathrm{4}}{a}+{C} \\ $$$${I}={I}\left(\mathrm{7}\right)−{I}\left(\mathrm{3}\right)=\left(\frac{\pi}{\mathrm{4}}.\mathrm{7}+{C}\right)−\left(\frac{\pi}{\mathrm{4}}.\mathrm{3}+{C}\right) \\ $$$${I}=\frac{\mathrm{7}\pi}{\mathrm{4}}−\frac{\mathrm{3}\pi}{\mathrm{4}}=\frac{\mathrm{4}\pi}{\mathrm{4}}=\pi \\ $$$$\:\:\mathrm{Final}\:\mathrm{Answer} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{7}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){lnx}}{dx}=\pi\: \\ $$$$ \\ $$

Commented by Spillover last updated on 01/May/25

$${thank}\:{you} \\ $$

Answered by maths2 last updated on 29/Apr/25

Let I=integral;x→(1/x) I=∫_0 ^∞ ((ln(1+x^7 )−ln(1+x^3 )−4ln(x))/((1+x^2 ).−ln(x)))dx =−I+4∫_0 ^∞ (dx/(1+x^2 ))⇒2I=4tan^(−1) (x)]_0 ^∞ =2π I=π

$${Let}\:{I}={integral};{x}\rightarrow\frac{\mathrm{1}}{{x}} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{7}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\mathrm{4}{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right).−{ln}\left({x}\right)}{dx} \\ $$$$\left.=−{I}+\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\Rightarrow\mathrm{2}{I}=\mathrm{4tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\infty} =\mathrm{2}\pi \\ $$$${I}=\pi \\ $$

Commented by Spillover last updated on 01/May/25