Question Number 100789 by 175 last updated on 28/Jun/20

Answered by maths mind last updated on 28/Jun/20

x^4 =u  ∫_0 ^1 ((x^7 −x^3 )/(ln(x)))dx=∫_0 ^1 ((x^3 (x^4 −1))/(ln(x)))dx  x^4 =y⇒x^3 dx=(dy/4)  ⇒∫_0 ^1 ((y−1)/(ln(y)))dy  −ln(y)=t⇒∫_0 ^(+∞) ((e^(−t) −1)/t)e^(−t)   =∫_0 ^(+∞) ((e^(−2t) −e^(−t) )/t)  ∫_0 ^(+∞) ∫_a ^b −e^(−xt) dx dt  =∫_0 ^(+∞) ((e^(−bt) −e^(−at) )/t)=∫_a ^b ∫_0 ^(+∞) −e^(−xt) dt.dx  .∫_a ^b ∫_0 ^(+∞) (e^(−xt) /x)dxdt=ln((a/b))  ∫_0 ^(+∞) ∫_1 ^2 −e^(−xt) dxdt=ln((1/2))=−ln(2)

Commented byAr Brandon last updated on 28/Jun/20

Hello Sir, I really admire the technics which you apply when solving maths questions especially when it concerns CALCULUS. Can you please recommend me any textbook amongst those which helped you improve your skills ? Any language. 😃

Commented bymaths mind last updated on 28/Jun/20

hello sir thank you  i didnt use realy used a book only som lesson in my sgudy  i used books i7 topologie and algebraic geometry  one Book that i find good realy chalanging problems  <impossible integral and sum>

Commented byAr Brandon last updated on 28/Jun/20

Thanks for your feedback. I'm currently using some JEE textbooks on CALCULUS, and I hope I'll step up.😃

Answered by abdomathmax last updated on 28/Jun/20

I =∫_0 ^1  ((x^7 −x^3 )/(lnx))dx  changement lnx =−t give  I =−∫_0 ^(+∞ )  ((e^(−7t) −e^(−3t) )/(−t))(−e^(−t) )dt  =−∫_0 ^∞   ((e^(−8t) −e^(−4t) )/t) dt  let f(a) =∫_0 ^∞  ((e^(−8t) −e^(−4t) )/t) e^(−at) dt  (a>0) we have f^′ (a) =−∫_0 ^∞ (e^(−8t) −e^(−4t) )e^(−at)  dt  =∫_0 ^∞ (e^(−(a+4)t) −e^(−(a+8)t) )dt  =[−(1/((a+4)))e^((a+4)t)  +(1/(a+8)) e^(−(a+8)t) ]_0 ^(+∞)   =(1/(a+4))−(1/(a+8)) ⇒f(a) =ln(((a+4)/(a+8))) +c(c=lim_(a→+∞) f(a)=0  ⇒f(a) =ln(((a+4)/(a+8)))f(0) =ln((1/2))=∫_0 ^∞  ((e^(−8t) −e^(−4t) )/t)dt  ⇒  I =−f(0) =ln2