# Value of $5050\cdot\int^1_0(1-x^{50})^{100} dx/\int^1_0(1-x^{50})^{101} dx$ [duplicate]

$$\frac{(5050)\int^1_0(1-x^{50})^{100} dx}{\int^1_0(1-x^{50})^{101} dx}$$ I tried solving both numerator and denominator via by-parts but it isn't giving me a conclusive solution. Any other suggestions?

## marked as duplicate by Ian Miller, Michael Hoppe, Nosrati, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 16 '17 at 5:55

Let's try integrating by parts and see what happens. $$\int_0^1 (1-x^n)^m \, dx = \left[ x(1-x^n)^m \right]_0^1 - \int_0^1 -nmx^n(1-x^n)^{m-1} \, dx = 0+ nm\int_0^1 x^n(1-x^n)^{m-1} \, dx.$$ We can fiddle with the right-hand side to get it into a more familiar form: $$\int_0^1 x^n(1-x^n)^{m-1} \, dx = \int_0^1 \left( 1 -(1-x^n) \right) (1-x^n)^{m-1} \, dx = \int_0^1 (1-x^n)^{m-1} \, dx - \int_0^1 (1-x^n)^{m} \, dx.$$ Collecting the copies of the integrals together, we find $$(1+nm)\int_0^1 (1-x^n)^m \, dx = nm \int_0^1 (1-x^n)^{m-1} \, dx,$$ or $$\frac{\int_0^1 (1-x^n)^{m-1} \, dx}{\int_0^1 (1-x^n)^m \, dx} = \frac{1+nm}{nm}.$$ Putting $n=50$, $m=101$, the right-hand side becomes $5051/5050$.

• What happened to $[ x(1-x^n)^m ]$ ? – A---B Apr 15 '17 at 1:09
• Yep, that should be evaluated at both endpoints and disappear. Fixed. Thanks. – Chappers Apr 15 '17 at 1:10
• very nice answer..@ Uddeshya Singh you can accept this answer because it answers fully your question – Marios Gretsas Apr 15 '17 at 1:13
• This answer is very clever.. – A---B Apr 15 '17 at 1:35

one way to solve it besides the change of variable,is by using the binomial theorem in both denominator and enumerator.

$\sum_{n=0}^{100}\binom{100}{n}(x^{50})^{100-n}(-1)^n=(1-x^{50})^{100}$

$\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Use the linearity of the integral and integrate each term of the series.

• I'm sorry but m not really well versed in usage of binomial theorem, Can u please guide further – The Dead Legend Apr 15 '17 at 0:51
• i edited ...if you want something else let me know – Marios Gretsas Apr 15 '17 at 0:55
• Got it. Done. answer is 5051? – The Dead Legend Apr 15 '17 at 0:56
• i dont think so.. – Marios Gretsas Apr 15 '17 at 1:01